Probability

1. Revision of probability


ID is: 3044 Seed is: 3935

Probability basics

Fill in the missing word or term for the sentence below:

Two events, C and D are _ if the outcomes of one event influences the outcome of the other.

NOTE: Remember that spelling is important! If you are unsure, look it up in your notes or textbook.

Answer:

Two events, C and D are if the outcomes of one event influences the outcome of the other.

one-of
type(string.nocase)
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For a list of terminology and definitions see the beginning of the Probability chapter in Everything Maths.

Two events, C and D are dependent if the outcomes of one event influences the outcome of the other.


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ID is: 3044 Seed is: 1730

Probability basics

Fill in the missing word or term for the sentence below:

The _ is the set of all possible outcomes of the experiment.

NOTE: Remember that spelling is important! If you are unsure, look it up in your notes or textbook.

Answer:

The is the set of all possible outcomes of the experiment.

one-of
type(string.nocase)
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STEP: <no title>
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For a list of terminology and definitions see the beginning of the Probability chapter in Everything Maths.

The sample space is the set of all possible outcomes of the experiment.


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Probability basics

Fill in the missing word or term for the sentence below:

A probability of _ means the outcome of the experiment will never be in the event set.

NOTE: Remember that spelling is important! If you are unsure, look it up in your notes or textbook.

Answer:

A probability of means the outcome of the experiment will never be in the event set.

one-of
type(string.nocase)
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STEP: <no title>
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For a list of terminology and definitions see the beginning of the Probability chapter in Everything Maths.

A probability of 0 means the outcome of the experiment will never be in the event set.


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Using the probability identity

Suppose we have two sets, A and B, which are in some sample space S. Furthermore, suppose we know the following probabilities about events from these sets:

P(A)=13P(B)=12P(A and B)=13

Use the probability identity to calculate P(A or B).

INSTRUCTION: Give an exact answer (do not round off).
Answer: P(A or B)=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the probability identity. If you do not remember it, you can find it here.


STEP: Substitute the values into the probablity identity and work out the answer
[−2 points ⇒ 0 / 2 points left]

This question is about the probability identity:

P(A or B)=P(A)+P(B)P(A and B)

This formula applies to any two sets inside of a sample space.

We have the three probabilities on the right hand side of the equation, so we just need to substitute in the values from the question and solve for P(A or B).

P(A or B)=(13)+(12)(13)=12

The answer is P(A or B)=12.


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ID is: 3138 Seed is: 1864

Using the probability identity

Suppose we have two sets, A and B, which are in some sample space S. Furthermore, suppose we know the following probabilities about events from these sets:

P(A)=14P(B)=712P(A and B)=14

Use the probability identity to calculate P(A or B).

INSTRUCTION: Give an exact answer (do not round off).
Answer: P(A or B)=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the probability identity. If you do not remember it, you can find it here.


STEP: Substitute the values into the probablity identity and work out the answer
[−2 points ⇒ 0 / 2 points left]

This question is about the probability identity:

P(A or B)=P(A)+P(B)P(A and B)

This formula applies to any two sets inside of a sample space.

We have the three probabilities on the right hand side of the equation, so we just need to substitute in the values from the question and solve for P(A or B).

P(A or B)=(14)+(712)(14)=712

The answer is P(A or B)=712.


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ID is: 3138 Seed is: 5520

Using the probability identity

Suppose we have two sets, K and L, which are in some sample space S. Furthermore, suppose we know the following probabilities about events from these sets:

P(K)=37P(L)=37P(K and L)=27

Use the probability identity to calculate P(K or L).

INSTRUCTION: Give an exact answer (do not round off).
Answer: P(K or L)=
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing down the probability identity. If you do not remember it, you can find it here.


STEP: Substitute the values into the probablity identity and work out the answer
[−2 points ⇒ 0 / 2 points left]

This question is about the probability identity:

P(A or B)=P(A)+P(B)P(A and B)

This formula applies to any two sets inside of a sample space.

We have the three probabilities on the right hand side of the equation, so we just need to substitute in the values from the question and solve for P(K or L).

P(K or L)=(37)+(37)(27)=47

The answer is P(K or L)=47.


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Exploring definitions

A student wants to understand the term "experiment". So the student stands looking at a busy highway. Which of the following is the most appropriate example of the term "experiment"?

Answer: The most appropriate choice is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look up information about this in your notes or a book.


STEP: Select the best option from the choices in the list
[−1 point ⇒ 0 / 1 points left]

We recall the definition of the term "experiment":

An experiment refers to an uncertain process

Therefore the most appropriate example of the term "experiment" is will a truck drive past? . The other choices are also outcomes but not related to this specific experiment, therefore they are incorrect.


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Exploring definitions

A student wants to understand the term "outcome". So the student rolls a die. Which of the following is the most appropriate example of the term "outcome"?

Answer: The most appropriate choice is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look up information about this in your notes or a book.


STEP: Select the best option from the choices in the list
[−1 point ⇒ 0 / 1 points left]

We recall the definition of the term "outcome":

An outcome of an experiment is a single result of that experiment

Therefore the most appropriate example of the term "outcome" is the die lands on the number 6 . The other choices are also outcomes but not related to this specific experiment, therefore they are incorrect.


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ID is: 1194 Seed is: 2034

Exploring definitions

A student wants to understand the term "experiment". So the student stands looking at a busy highway. Which of the following is the most appropriate example of the term "experiment"?

Answer: The most appropriate choice is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look up information about this in your notes or a book.


STEP: Select the best option from the choices in the list
[−1 point ⇒ 0 / 1 points left]

We recall the definition of the term "experiment":

An experiment refers to an uncertain process

Therefore the most appropriate example of the term "experiment" is will there be an accident? . The other choices are also outcomes but not related to this specific experiment, therefore they are incorrect.


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ID is: 3135 Seed is: 9325

Relative frequency compared to theoretical probability

Tshegofatso just flipped a coin 3 times. The results of the flips were:

Flip 1 Flip 2 Flip 3 Next Flip
Tails Heads Heads ?

Answer the three questions which follow about this situation.

  1. What is the most likely outcome of her next flip?

    Answer: For the next flip, .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The results of the first 3 flips do not influence the next flip.


    STEP: Use the theoretical probability to determine the answer
    [−1 point ⇒ 0 / 1 points left]

    In this question, Tshegofatso is flipping a coin. She already flipped the coin 3 times. From the list of results we can see that she got heads 2 times and tails 1 time. The question asks us what is likely to happen in the next flip.

    The key is to realise that the past outcomes do not change the theoretical probability of the next flip. The theoretical probability for heads and tails is 12 and that does not change based on earlier outcomes. Since there are more heads than tails in the first 3 flips, it is tempting to think that the next flip is likely to be tails. But that is not true. Even if someone flips a coin and gets heads 10 times in a row, the theoretical probability is still 12 that the next flip will be heads again.

    The correct choice from the list is: both outcomes are equally likely.


    Submit your answer as:
  2. What is the relative frequency of tails for the first 3 flips? Give your answer as a fraction (not a percentage or a decimal).

    Answer: The relative frequency of tails is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Relative frequency is the number of times a specific outcome happened divided by the total number of trials. You need to do that calculation using the information about the coin flips.


    STEP: Calculate the relative frequency based on the results of the flips
    [−1 point ⇒ 0 / 1 points left]

    We need to calculate the relative frequency of tails for the first 3 flips. Relative frequency is the ratio of how many times something happens to the total number of times we tried the experiment. In this case, that means the ratio of how many times Tshegofatso got tails to the total number of flips. This is:

    Relative frequency of tails =number of tailsnumber of flips=13

    The relative frequency of tails from the first 3 flips is 13.


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  3. The theoretical probability for getting tails for any flip is 12. However, the relative frequency in Question 2 was not equal to 12. How is this possible? Select your answer from the choices below.

    A The coin must be an unfair one.
    B Tshegofatso can not flip coins properly.
    C The theoretical probability must be wrong.
    D Relative frequency often does not equal theoretical probability.
    E None of the above
    Answer: It is possible because .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Theoretical probability applies to each and every individual coin flip. The relative frequency summarises the outcome of an experiment. Must they always be the same?


    STEP: Distinguish between relative frequency and theoretical probability
    [−1 point ⇒ 0 / 1 points left]

    This question is about the relationship between theoretical probability and relative frequency. The relative frequency depends on the sequence of outcomes that we observe while doing an experiment like flipping coins many times. The relative frequency can be different every time we redo the experiment. This is actually what probability means: the outcome is not predictable. When flipping a coin, it might land on heads and it might land on tails.

    Theoretical probability tells us how likely the different outcomes are for a single event. For a coin flip, getting heads or tails are equally likely results. So the probability for heads and tails is 12 each.

    However, each flip is independent of the flips before and after it. As a result, it is common that the relative frequency of an event does not exactly equal the theoretical probability of the event. But the more times we flip the coin, the closer and closer the relative frequency gets to the theoretical probability.

    The correct statement from the list is: Relative frequency often does not equal theoretical probability. So the correct choice in the table is D.


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ID is: 3135 Seed is: 1347

Relative frequency compared to theoretical probability

Lidija just flipped a coin 3 times. The results of the flips were:

Flip 1 Flip 2 Flip 3 Next Flip
Tails Tails Heads ?

Answer the three questions which follow about this situation.

  1. What is the most likely outcome of her next flip?

    Answer: For the next flip, .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The results of the first 3 flips do not influence the next flip.


    STEP: Use the theoretical probability to determine the answer
    [−1 point ⇒ 0 / 1 points left]

    In this question, Lidija is flipping a coin. She already flipped the coin 3 times. From the list of results we can see that she got heads 1 time and tails 2 times. The question asks us what is likely to happen in the next flip.

    The key is to realise that the past outcomes do not change the theoretical probability of the next flip. The theoretical probability for heads and tails is 12 and that does not change based on earlier outcomes. Since there are more tails than heads in the first 3 flips, it is tempting to think that the next flip is likely to be heads. But that is not true. Even if someone flips a coin and gets tails 10 times in a row, the theoretical probability is still 12 that the next flip will be tails again.

    The correct choice from the list is: both outcomes are equally likely.


    Submit your answer as:
  2. What is the relative frequency of heads for the first 3 flips? Give your answer as a fraction (not a percentage or a decimal).

    Answer: The relative frequency of heads is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Relative frequency is the number of times a specific outcome happened divided by the total number of trials. You need to do that calculation using the information about the coin flips.


    STEP: Calculate the relative frequency based on the results of the flips
    [−1 point ⇒ 0 / 1 points left]

    We need to calculate the relative frequency of heads for the first 3 flips. Relative frequency is the ratio of how many times something happens to the total number of times we tried the experiment. In this case, that means the ratio of how many times Lidija got heads to the total number of flips. This is:

    Relative frequency of heads =number of headsnumber of flips=13

    The relative frequency of heads from the first 3 flips is 13.


    Submit your answer as:
  3. The theoretical probability for getting heads for any flip is 12. However, the relative frequency in Question 2 was not equal to 12. What will happen to the relative frequency if Lidija keeps flipping the coin? Select your answer from the choices below.

    A Lidija might miss the bus home.
    B The pattern of more tails than heads will continue.
    C The relative frequency will get closer and closer to one half.
    D The relative frequency will get closer and closer to 0.
    E None of the above
    Answer: If she keeps flipping, .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The more times Lidija flips the coin, the more you can expect the number of heads and tails to balance. If that happens, what will happen to the relative frequency?


    STEP: Distinguish between relative frequency and theoretical probability
    [−1 point ⇒ 0 / 1 points left]

    This question is about the relationship between theoretical probability and relative frequency. The relative frequency depends on the sequence of outcomes that we observe while doing an experiment like flipping coins many times. The relative frequency can be different every time we redo the experiment. This is actually what probability means: the outcome is not predictable. When flipping a coin, it might land on heads and it might land on tails.

    Theoretical probability tells us how likely the different outcomes are for a single event. For a coin flip, getting heads or tails are equally likely results. So the probability for heads and tails is 12 each.

    However, each flip is independent of the flips before and after it. As a result, it is common that the relative frequency of an event does not exactly equal the theoretical probability of the event. But the more times we flip the coin, the closer and closer the relative frequency gets to the theoretical probability.

    The correct statement from the list is: The relative frequency will get closer and closer to one half. So the correct choice in the table is C.


    Submit your answer as:

ID is: 3135 Seed is: 1305

Relative frequency compared to theoretical probability

Sindisiwe just flipped a coin 4 times. The results of the flips were:

Flip 1 Flip 2 Flip 3 Flip 4 Next Flip
Tails Heads Heads Heads ?

Answer the three questions which follow about this situation.

  1. What is the most likely outcome of her next flip?

    Answer: For the next flip, .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The results of the first 4 flips do not influence the next flip.


    STEP: Use the theoretical probability to determine the answer
    [−1 point ⇒ 0 / 1 points left]

    In this question, Sindisiwe is flipping a coin. She already flipped the coin 4 times. From the list of results we can see that she got heads 3 times and tails 1 time. The question asks us what is likely to happen in the next flip.

    The key is to realise that the past outcomes do not change the theoretical probability of the next flip. The theoretical probability for heads and tails is 12 and that does not change based on earlier outcomes. Since there are more heads than tails in the first 4 flips, it is tempting to think that the next flip is likely to be tails. But that is not true. Even if someone flips a coin and gets heads 10 times in a row, the theoretical probability is still 12 that the next flip will be heads again.

    The correct choice from the list is: both outcomes are equally likely.


    Submit your answer as:
  2. What is the relative frequency of heads for the first 4 flips? Give your answer as a fraction (not a percentage or a decimal).

    Answer: The relative frequency of heads is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Relative frequency is the number of times a specific outcome happened divided by the total number of trials. You need to do that calculation using the information about the coin flips.


    STEP: Calculate the relative frequency based on the results of the flips
    [−1 point ⇒ 0 / 1 points left]

    We need to calculate the relative frequency of heads for the first 4 flips. Relative frequency is the ratio of how many times something happens to the total number of times we tried the experiment. In this case, that means the ratio of how many times Sindisiwe got heads to the total number of flips. This is:

    Relative frequency of heads =number of headsnumber of flips=34

    The relative frequency of heads from the first 4 flips is 34.


    Submit your answer as:
  3. The theoretical probability for getting heads for any flip is 12. However, the relative frequency in Question 2 was not equal to 12. What will happen to the relative frequency if Sindisiwe keeps flipping the coin? Select your answer from the choices below.

    A The relative frequency will get closer and closer to one half.
    B The pattern of more heads than tails will continue.
    C The relative frequency will get closer and closer to 1.
    D She will probably lose her coin.
    E None of the above
    Answer: If she keeps flipping, .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The more times Sindisiwe flips the coin, the more you can expect the number of heads and tails to balance. If that happens, what will happen to the relative frequency?


    STEP: Distinguish between relative frequency and theoretical probability
    [−1 point ⇒ 0 / 1 points left]

    This question is about the relationship between theoretical probability and relative frequency. The relative frequency depends on the sequence of outcomes that we observe while doing an experiment like flipping coins many times. The relative frequency can be different every time we redo the experiment. This is actually what probability means: the outcome is not predictable. When flipping a coin, it might land on heads and it might land on tails.

    Theoretical probability tells us how likely the different outcomes are for a single event. For a coin flip, getting heads or tails are equally likely results. So the probability for heads and tails is 12 each.

    However, each flip is independent of the flips before and after it. As a result, it is common that the relative frequency of an event does not exactly equal the theoretical probability of the event. But the more times we flip the coin, the closer and closer the relative frequency gets to the theoretical probability.

    The correct statement from the list is: The relative frequency will get closer and closer to one half. So the correct choice in the table is A.


    Submit your answer as:

ID is: 1188 Seed is: 2041

Theoretical probability

A student finds a 6 sided die and then rolls the die once on a table. What is the probability that the die lands on either 4 or 5?

INSTRUCTION: Write your answer as a decimal (correct to 2 decimal places).
Answer: The probability is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must use the formula for probability:

P(E)=n(E)n(S)

STEP: Use the formula for theoretical probability
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

P(E)=n(E)n(S)

Secondly we identify the values for the situation described in the question:

n(E)=number of outcomes in the event set=2n(S)=number of possible outcomes in the sample space=6

Finally, we substitute into the formula and work out the answer.

P(E)=n(E)n(S)=26P(E)0.3333

Remember that the question says your answer should be "correct to 2 decimal places," which means rounding to two places. Therefore, the probability that the die lands on either 4 or 5 0.33.


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ID is: 1188 Seed is: 3875

Theoretical probability

A student finds a deck of 52 cards and then takes one card from the deck. What is the probability that the card's suit is Spades?

INSTRUCTION: Write your answer as a simplified fraction.
Answer: The probability is .
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must use the formula for probability:

P(E)=n(E)n(S)

STEP: Use the formula for theoretical probability
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

P(E)=n(E)n(S)

Secondly we identify the values for the situation described in the question:

n(E)=number of outcomes in the event set=13n(S)=number of possible outcomes in the sample space=52

Finally, we substitute into the formula and work out the answer.

P(E)=n(E)n(S)=1352P(E)=14

Therefore, the probability that the card's suit is Spades =14.


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ID is: 1188 Seed is: 5095

Theoretical probability

A student finds a 6 sided die and then rolls the die once on a table. What is the probability that the die lands on either 4 or 5?

INSTRUCTION: Write your answer as a percentage (rounded to the nearest whole number).
Answer: The probability is %.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You must use the formula for probability:

P(E)=n(E)n(S)

STEP: Use the formula for theoretical probability
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

P(E)=n(E)n(S)

Secondly we identify the values for the situation described in the question:

n(E)=number of outcomes in the event set=2n(S)=number of possible outcomes in the sample space=6

Finally, we substitute into the formula and work out the answer.

P(E)=n(E)n(S)=26P(E)33.33%

Remember that the question says your answer should be "rounded to the nearest whole number." Therefore, the probability that the die lands on either 4 or 5 33%.


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ID is: 3137 Seed is: 2106

Applying theoretical probability to experimental trials

The theoretical probability for a special outcome is 14. Someone does an experiment about this event with 16 trials.

  1. Out of the 16 trials, how many times should we expect to get the special outcome? In other words, for the 16 trials, what is the most likely number of times we will get the special outcome?

    Answer: We expect the special outcome to happen times.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The theoretical probability tells you how frequently you can expect to get the special outcome. So use the theoretical probability to get the answer.


    STEP: Use the theoretical probability to get the answer
    [−1 point ⇒ 0 / 1 points left]

    This question tells us that the theoretical probability for some special outcome is 14. This means that we should expect the outcome to happen 1 time out of every 4 trials. For example, suppose the "special outcome" is picking a red marble from a sack of marbles. If we take a marble out of the sack 4 times (putting the marble back each time), we will expect to get a red marble 1 time.

    For this question, there are 16 trials. We expect 14 of the trials to result in the special outcome:

    special outcomes =(probability of the outcome)(number of trials)=1416=4

    Note: this calculation is related to the theoretical probability formula:

    P(E)=n(E)n(S)

    P(E) is the theoretical probability. n(S) is the number of trials (the total number of events in the sample space, no matter what the outcome is). n(E) is the number of outcomes in the event set, which are the special outcomes:

    14=special outcomes number of trials14(number of trials)=special outcomes 

    This is the same as the calculation we did above.

    In the experiment with 16 trials, we expect to get the special outcome 4 times.


    Submit your answer as:
  2. In reality, the number of trials with the special outcome might not be the number from Question 1: it could be more or less. This is because the relative frequency does not always match the theoretical probability. For example, any of these four results is possible:

    3 special outcomes from 16 trials10 special outcomes from 16 trials15 special outcomes from 16 trials16 special outcomes from 16 trials

    Given that the theoretical probability is still 14, which of the four results listed above is the most likely?

    Answer: The most likely result from the list is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The theoretical probability tells us that the most likely number of special outcomes is 4. The further you get from that number of special outcomes, the less likely it is.


    STEP: Compare the choices in the list to the result of Question 1
    [−1 point ⇒ 0 / 1 points left]

    The question now asks us to identify the most likely number of special outcomes from a list of choices. We can answer this using the concept of theoretical probability. The theoretical probability tells us what the most likely number of special outcomes is. In Question 1 we found that this is 4 of the 16 trials. But of course the experiment might not result in 4 special outcomes. That is because the outcomes are not predictable; they are probabilistic. For each individual trial, we cannot know if we will get the special outcome or not until we do the trial and see what we get.

    So 4 special outcomes is the most likely result of the experiment. But it is possible, and a little less likely, that we might get 3 or 5 special outcomes. It is also possible, though even less likely, that we might get 2 or 6 special outcomes. In fact, the further we get from 4 special outcomes, the less likely that result becomes. This makes sense: if we expect 4 special outcomes, it is quite unlikely that we will get the special outcome for all 16 trials.

    Now back to our list of choices. We need to identify the most likely number of special outcomes from a list of choices. The most likely option is the choice closest to the theoretically probable number of special outcomes.

    So the most likely choice from the list is: 3 special outcomes for 16 trials.


    Submit your answer as:

ID is: 3137 Seed is: 657

Applying theoretical probability to experimental trials

The theoretical probability for a special outcome is 15. Someone does an experiment about this event with 15 trials.

  1. Out of the 15 trials, how many times should we expect to get the special outcome? In other words, for the 15 trials, what is the most likely number of times we will get the special outcome?

    Answer: We expect the special outcome to happen times.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The theoretical probability tells you how frequently you can expect to get the special outcome. So use the theoretical probability to get the answer.


    STEP: Use the theoretical probability to get the answer
    [−1 point ⇒ 0 / 1 points left]

    This question tells us that the theoretical probability for some special outcome is 15. This means that we should expect the outcome to happen 1 time out of every 5 trials. For example, suppose the "special outcome" is picking a red marble from a sack of marbles. If we take a marble out of the sack 5 times (putting the marble back each time), we will expect to get a red marble 1 time.

    For this question, there are 15 trials. We expect 15 of the trials to result in the special outcome:

    special outcomes =(probability of the outcome)(number of trials)=1515=3

    Note: this calculation is related to the theoretical probability formula:

    P(E)=n(E)n(S)

    P(E) is the theoretical probability. n(S) is the number of trials (the total number of events in the sample space, no matter what the outcome is). n(E) is the number of outcomes in the event set, which are the special outcomes:

    15=special outcomes number of trials15(number of trials)=special outcomes 

    This is the same as the calculation we did above.

    In the experiment with 15 trials, we expect to get the special outcome 3 times.


    Submit your answer as:
  2. In reality, the number of trials with the special outcome might not be the number from Question 1: it could be more or less. This is because the relative frequency does not always match the theoretical probability. For example, any of these four results is possible:

    4 special outcomes from 15 trials11 special outcomes from 15 trials14 special outcomes from 15 trials15 special outcomes from 15 trials

    Given that the theoretical probability is still 15, which of the four results listed above is the most likely?

    Answer: The most likely result from the list is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The theoretical probability tells us that the most likely number of special outcomes is 3. The further you get from that number of special outcomes, the less likely it is.


    STEP: Compare the choices in the list to the result of Question 1
    [−1 point ⇒ 0 / 1 points left]

    The question now asks us to identify the most likely number of special outcomes from a list of choices. We can answer this using the concept of theoretical probability. The theoretical probability tells us what the most likely number of special outcomes is. In Question 1 we found that this is 3 of the 15 trials. But of course the experiment might not result in 3 special outcomes. That is because the outcomes are not predictable; they are probabilistic. For each individual trial, we cannot know if we will get the special outcome or not until we do the trial and see what we get.

    So 3 special outcomes is the most likely result of the experiment. But it is possible, and a little less likely, that we might get 2 or 4 special outcomes. It is also possible, though even less likely, that we might get 1 or 5 special outcomes. In fact, the further we get from 3 special outcomes, the less likely that result becomes. This makes sense: if we expect 3 special outcomes, it is quite unlikely that we will get the special outcome for all 15 trials.

    Now back to our list of choices. We need to identify the most likely number of special outcomes from a list of choices. The most likely option is the choice closest to the theoretically probable number of special outcomes.

    So the most likely choice from the list is: 4 special outcomes for 15 trials.


    Submit your answer as:

ID is: 3137 Seed is: 9749

Applying theoretical probability to experimental trials

The theoretical probability for a special outcome is 34. Someone does an experiment about this event with 16 trials.

  1. Out of the 16 trials, how many times should we expect to get the special outcome? In other words, for the 16 trials, what is the most likely number of times we will get the special outcome?

    Answer: We expect the special outcome to happen times.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The theoretical probability tells you how frequently you can expect to get the special outcome. So use the theoretical probability to get the answer.


    STEP: Use the theoretical probability to get the answer
    [−1 point ⇒ 0 / 1 points left]

    This question tells us that the theoretical probability for some special outcome is 34. This means that we should expect the outcome to happen 3 times out of every 4 trials. For example, suppose the "special outcome" is picking a red marble from a sack of marbles. If we take a marble out of the sack 4 times (putting the marble back each time), we will expect to get a red marble 3 times.

    For this question, there are 16 trials. We expect 34 of the trials to result in the special outcome:

    special outcomes =(probability of the outcome)(number of trials)=3416=12

    Note: this calculation is related to the theoretical probability formula:

    P(E)=n(E)n(S)

    P(E) is the theoretical probability. n(S) is the number of trials (the total number of events in the sample space, no matter what the outcome is). n(E) is the number of outcomes in the event set, which are the special outcomes:

    34=special outcomes number of trials34(number of trials)=special outcomes 

    This is the same as the calculation we did above.

    In the experiment with 16 trials, we expect to get the special outcome 12 times.


    Submit your answer as:
  2. In reality, the number of trials with the special outcome might not be the number from Question 1: it could be more or less. This is because the relative frequency does not always match the theoretical probability. For example, any of these four results is possible:

    1 special outcome from 16 trials3 special outcomes from 16 trials6 special outcomes from 16 trials13 special outcomes from 16 trials

    Given that the theoretical probability is still 34, which of the four results listed above is the most likely?

    Answer: The most likely result from the list is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The theoretical probability tells us that the most likely number of special outcomes is 12. The further you get from that number of special outcomes, the less likely it is.


    STEP: Compare the choices in the list to the result of Question 1
    [−1 point ⇒ 0 / 1 points left]

    The question now asks us to identify the most likely number of special outcomes from a list of choices. We can answer this using the concept of theoretical probability. The theoretical probability tells us what the most likely number of special outcomes is. In Question 1 we found that this is 12 of the 16 trials. But of course the experiment might not result in 12 special outcomes. That is because the outcomes are not predictable; they are probabilistic. For each individual trial, we cannot know if we will get the special outcome or not until we do the trial and see what we get.

    So 12 special outcomes is the most likely result of the experiment. But it is possible, and a little less likely, that we might get 11 or 13 special outcomes. It is also possible, though even less likely, that we might get 10 or 14 special outcomes. In fact, the further we get from 12 special outcomes, the less likely that result becomes. This makes sense: if we expect 12 special outcomes, it is quite unlikely that we will never get the special outcome at all (for none of the trials).

    Now back to our list of choices. We need to identify the most likely number of special outcomes from a list of choices. The most likely option is the choice closest to the theoretically probable number of special outcomes.

    So the most likely choice from the list is: 13 special outcomes for 16 trials.


    Submit your answer as:

ID is: 1638 Seed is: 4104

Revision: calculations with probability

In a group of 277 students 130 take Natural Science and 153 take Mathematics while 38 chose neither subject.

If a student is selected at random, what is the probability that the student does not take Natural Science or Mathematics?

Give your answer as a number

INSTRUCTION: Round your answer to one decimal place.
Answer: The probability is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
Construct a Venn Diagram which represents this information
STEP: Shade the relevant areas on the Venn diagram
[−2 points ⇒ 0 / 2 points left]

You can find the answer from the given information, how many students took neither subject?

The Venn diagram represents the event.

NOTE: Remember: ("and" is the same as intersection ) and ("or" is the same as union )

n(N or M)=n(N)+n(M)(N and M)
P( not N or M)=(38277)

The final answer is: P=38277=0.1.


Submit your answer as:

ID is: 1638 Seed is: 6403

Revision: calculations with probability

In a group of 277 students 149 take Natural Science and 135 take Afrikaans while 27 chose neither subject.

If a student is selected at random, what is the probability that the student takes Afrikaans only?

INSTRUCTION: Write your answer as a simplified fraction.
Answer: The probability is .
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
Construct a Venn Diagram which represents this information
STEP: Shade the relevant areas on the Venn diagram
[−2 points ⇒ 0 / 2 points left]

This question refers to the Probabiltiy of Afrikaans written as P(Afrikaans not Natural Science).

The Venn diagram represents the event.

NOTE: Remember: ("and" is the same as intersection ) and ("or" is the same as union )

The Venn diagram is divided into 4 sections. The shaded area represents P(Afrikaans not Natural Science). First complete all the given information on the diagram, then solve the missing parts. We follow these steps:

Step 1:
Union = total students - students that took neither subject
Union = (277) - (27) = 250

Step 2: Union = (Natural Science) + (Afrikaans) - intersection
Rearrange the formula:
Intersection = (Natural Science) + (Afrikaans) - union
Intersection = 34

Step 3: All of Afrikaans - intersection = Area of (Afrikaans not Natural Science)
(Afrikaans not Natural Science) = 101

P(A not N)=(101277)

The final answer is: P=101277.


Submit your answer as:

ID is: 1638 Seed is: 9166

Revision: calculations with probability

In a group of 266 students 111 take Natural Science and 168 take Mathematics while 36 chose neither subject.

If a student is selected at random, what is the probability that the student takes both Natural Science and Mathematics?

Give your answer as a number

INSTRUCTION: Round your answer to one decimal place.
Answer: The probability is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Construct a Venn Diagram which represents this information
STEP: Shade the relevant areas on the Venn diagram
[−3 points ⇒ 0 / 3 points left]

The question refers to the INTERSECTON of Natural Science and Mathematics.

The Venn diagram represents the event.

NOTE: Remember: ("and" is the same as intersection ) and ("or" is the same as union )

The Venn diagram is divided into 4 sections. Fill in the given information and calculate the missing section which is represented by the shaded area. In this case you want the amount of students taking N M. We follow these steps:

Step 1:
Union = total students - students that took neither subject
Union = (266) - (36) = 230

Step 2: Union = (Natural Science) + (Mathematics) - intersection
Rearrange the formula:
Intersection = (Natural Science) + (Mathematics) - union
Intersection = 49

n(N or M)=n(N)+n(M)n(N and M)
n(NM)=n(N)+n(M)n(NM)
n(NM)=n(N)+n(M)n(NM)
(NM)=111+168230
P(NM)=49266

The final answer is: P=738=0.2.


Submit your answer as:

ID is: 1189 Seed is: 8013

Relative frequency

A die is tossed 35 times and lands 4 times on the number 5.

What is the relative frequency of observing the die land on the number 5? Write your answer correct to 2 decimal places.

Answer: The relative frequency is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by writing down the formula for the relative frequency, f=pt. Then figure out where to substitute the numbers in the question.


STEP: Use the values given with the formula for the relative frequency
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

f=pt

Secondly we identify variables needed:

p=number of positive trials=4t=total number of trials=35

Finally, we solve the equation:

f=pt=435=0.11

Therefore, the relative frequency of observing the die on the number 5 is 0.11.


Submit your answer as:

ID is: 1189 Seed is: 1055

Relative frequency

A die is tossed 26 times and lands 4 times on the number 3.

What is the relative frequency of observing the die land on the number 3? Write your answer correct to 2 decimal places.

Answer: The relative frequency is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by writing down the formula for the relative frequency, f=pt. Then figure out where to substitute the numbers in the question.


STEP: Use the values given with the formula for the relative frequency
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

f=pt

Secondly we identify variables needed:

p=number of positive trials=4t=total number of trials=26

Finally, we solve the equation:

f=pt=426=0.15

Therefore, the relative frequency of observing the die on the number 3 is 0.15.


Submit your answer as:

ID is: 1189 Seed is: 9927

Relative frequency

A coin is tossed 41 times and lands 20 times on tails.

What is the relative frequency of observing the coin land on tails? Write your answer correct to 2 decimal places.

Answer: The relative frequency is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by writing down the formula for the relative frequency, f=pt. Then figure out where to substitute the numbers in the question.


STEP: Use the values given with the formula for the relative frequency
[−1 point ⇒ 0 / 1 points left]

To calculate probability, we first recall the formula:

f=pt

Secondly we identify variables needed:

p=number of positive trials=20t=total number of trials=41

Finally, we solve the equation:

f=pt=2041=0.49

Therefore, the relative frequency of observing the coin on tails is 0.49.


Submit your answer as:

ID is: 1187 Seed is: 19

Probability identities

A group of students is given the following event sets:

Event Set B12356
Event Set (A or B)12356
Event Set (A and B)2356

The sample space is {1;2;3;4;5;6}.

They are asked to calculate the value of P(A). They get stuck, and you offer to calculate it for them.

Answer: P(A) =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing down the probability identity. If you are not sure what that is, you can find a review here.


STEP: Rearrange the probabilty identity for the value we want
[−1 point ⇒ 2 / 3 points left]

To calculate probability, we use the probability identity:

P(A or B)=P(A)+P(B)P(A and B)

Rearranging the formula to make P(A) the subject, we get:

P(A)=P(A or B)P(B)+P(A and B)

STEP: Calculate each probability that we need
[−1 point ⇒ 1 / 3 points left]

Now we need to identify variables needed:

P(B)=n(B)n(S)=56P(A or B)=n(A or B)n(S)=56P(A and B)=n(A and B)n(S)=46

STEP: Substitute in the values we know and evaluate
[−1 point ⇒ 0 / 3 points left]

Finally, substitute in and solve for the unknown value P(A):

P(A)=P(A or B)P(B)+P(A and B)P(A)=(56)(56)+(46)P(A)=23

Therefore, the value of P(A) is 23.


Submit your answer as:

ID is: 1187 Seed is: 1499

Probability identities

A group of students is given the following event sets:

Event Set B12346
Event Set (A or B)12346
Event Set (A and B)23

The sample space is {1;2;3;4;5;6}.

They are asked to calculate the value of P(A). They get stuck, and you offer to calculate it for them.

Answer: P(A) =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing down the probability identity. If you are not sure what that is, you can find a review here.


STEP: Rearrange the probabilty identity for the value we want
[−1 point ⇒ 2 / 3 points left]

To calculate probability, we use the probability identity:

P(A or B)=P(A)+P(B)P(A and B)

Rearranging the formula to make P(A) the subject, we get:

P(A)=P(A or B)P(B)+P(A and B)

STEP: Calculate each probability that we need
[−1 point ⇒ 1 / 3 points left]

Now we need to identify variables needed:

P(B)=n(B)n(S)=56P(A or B)=n(A or B)n(S)=56P(A and B)=n(A and B)n(S)=26

STEP: Substitute in the values we know and evaluate
[−1 point ⇒ 0 / 3 points left]

Finally, substitute in and solve for the unknown value P(A):

P(A)=P(A or B)P(B)+P(A and B)P(A)=(56)(56)+(26)P(A)=13

Therefore, the value of P(A) is 13.


Submit your answer as:

ID is: 1187 Seed is: 6509

Probability identities

A group of students is given the following event sets:

Event Set B5
Event Set (A or B)156
Event Set (A and B)empty

The sample space is {1;2;3;4;5;6}.

They are asked to calculate the value of P(A). They get stuck, and you offer to calculate it for them.

Answer: P(A) =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing down the probability identity. If you are not sure what that is, you can find a review here.


STEP: Rearrange the probabilty identity for the value we want
[−1 point ⇒ 2 / 3 points left]

To calculate probability, we use the probability identity:

P(A or B)=P(A)+P(B)P(A and B)

Rearranging the formula to make P(A) the subject, we get:

P(A)=P(A or B)P(B)+P(A and B)

STEP: Calculate each probability that we need
[−1 point ⇒ 1 / 3 points left]

Now we need to identify variables needed:

P(B)=n(B)n(S)=16P(A or B)=n(A or B)n(S)=36P(A and B)=n(A and B)n(S)=06

STEP: Substitute in the values we know and evaluate
[−1 point ⇒ 0 / 3 points left]

Finally, substitute in and solve for the unknown value P(A):

P(A)=P(A or B)P(B)+P(A and B)P(A)=(36)(16)+(06)P(A)=13

Therefore, the value of P(A) is 13.


Submit your answer as:

2. Equiprobable sample space

3. Mutually exclusive events


ID is: 1191 Seed is: 4075

Mutually exclusive events

Consider the following Venn diagram, which shows two sets in the sample space {n:n ϵ Z, 1n15} (the first 15 natural numbers).

Are the following two sets mutually exclusive, or not?

Set 1:Set 2:(A or B)(A or B)
Answer: Are (A or B) and (A or B) mutually exclusive?
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the two sets (A or B) and (A or B) on the Venn diagram.


STEP: Identify the elements in the two sets and compare them
[−1 point ⇒ 0 / 1 points left]

We need to determine whether or not two sets are mutually exclusive. Two sets are called mutually exclusive if they do not share any elements (outcomes). To find the answer, we need to compare the elements in the two sets.

  • The event set for (A or B) is: {1;2;3;4;6;7;8;11;12;14;15}.
  • The event set for (A or B) is: {5;9;10;13}.

The question we must ask: do the sets have any elements in common? By comparing the sets, we can identify the following overlapping event set: {empty}.

Therefore the event sets (A or B) and (A or B) are mutually exclusive in this example. The correct answer is Yes.


Submit your answer as:

ID is: 1191 Seed is: 1085

Mutually exclusive events

Consider the following Venn diagram, which shows two sets in the sample space {n:n ϵ Z, 1n15} (the first 15 natural numbers).

Are the following two sets mutually exclusive, or not?

Set 1:Set 2:(A and B)(A or B)
Answer: Are (A and B) and (A or B) mutually exclusive?
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the two sets (A and B) and (A or B) on the Venn diagram.


STEP: Identify the elements in the two sets and compare them
[−1 point ⇒ 0 / 1 points left]

We need to determine whether or not two sets are mutually exclusive. Two sets are called mutually exclusive if they do not share any elements (outcomes). To find the answer, we need to compare the elements in the two sets.

  • The event set for (A and B) is: {2;7;9;13;15}.
  • The event set for (A or B) is: {5;8;14}.

The question we must ask: do the sets have any elements in common? By comparing the sets, we can identify the following overlapping event set: {empty}.

Therefore the event sets (A and B) and (A or B) are mutually exclusive in this example. The correct answer is Yes.


Submit your answer as:

ID is: 1191 Seed is: 482

Mutually exclusive events

Consider the following Venn diagram, which shows two sets in the sample space {n:n ϵ Z, 1n15} (the first 15 natural numbers).

Are the following two sets mutually exclusive, or not?

Set 1:Set 2:(A or B)(A or B)
Answer: Are (A or B) and (A or B) mutually exclusive?
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Identify the two sets (A or B) and (A or B) on the Venn diagram.


STEP: Identify the elements in the two sets and compare them
[−1 point ⇒ 0 / 1 points left]

We need to determine whether or not two sets are mutually exclusive. Two sets are called mutually exclusive if they do not share any elements (outcomes). To find the answer, we need to compare the elements in the two sets.

  • The event set for (A or B) is: {}.
  • The event set for (A or B) is: {1;2;3;4;5;6;7;8;9;10;11;12;13;14;15}.

The question we must ask: do the sets have any elements in common? By comparing the sets, we can identify the following overlapping event set: {empty}.

Therefore the event sets (A or B) and (A or B) are mutually exclusive in this example. The correct answer is Yes.


Submit your answer as:

ID is: 3166 Seed is: 7591

Mutually exclusive sets

You are given the following two sets:

A={46;107;98;105}B={105;109;112;53}

From the drop down box below choose whether the values in these sets are mutually exclusive or not.

Answer: These sets are .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Mutually exclusive sets can never have common values.


STEP: Compare the values in the two given sets
[−1 point ⇒ 0 / 1 points left]

When sets are not mutually exclusive that means at least one element belongs to both sets at the same time. In this specific case an element that belongs in both sets is: 105.

The correct choice from the list is: not mutually exclusive.


Submit your answer as:

ID is: 3166 Seed is: 8862

Mutually exclusive sets

You are given the following two sets:

A={79;97;109;93}B={103;49;84;96}

From the drop down box below choose whether the values in these sets are mutually exclusive or not.

Answer: These sets are .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Mutually exclusive sets can never have common values.


STEP: Compare the values in the two given sets
[−1 point ⇒ 0 / 1 points left]

Mutually exclusive sets mean that we cannot pick one element which belongs to both of the sets. So whatever number we pick, it cannot belong to both sets. If we compare the elements in these two sets, we find that not a single value belongs in both sets. This is what the diagram represents below.

The correct choice from the list is: mutually exclusive.


Submit your answer as:

ID is: 3166 Seed is: 2773

Mutually exclusive sets

You are given the following two sets:

A={101;94;103;112}B={113;54;58;97}

From the drop down box below choose whether the values in these sets are mutually exclusive or not.

Answer: These sets are .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Mutually exclusive sets can never have common values.


STEP: Compare the values in the two given sets
[−1 point ⇒ 0 / 1 points left]

Mutually exclusive sets mean that we cannot pick one element which belongs to both of the sets. So whatever number we pick, it cannot belong to both sets. If we compare the elements in these two sets, we find that not a single value belongs in both sets. This is what the diagram represents below.

The correct choice from the list is: mutually exclusive.


Submit your answer as:

ID is: 4344 Seed is: 5403

Prove that two events are mutually exclusive

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

Given two events A and B:

  • P(A)=0.4
  • P(B)=0.5
  • P(A or B)=0.8

Are the events A and B mutually exclusive? Justify your answer with appropriate calculations.

INSTRUCTION: Choose the most correct option to answer this question.
Answer:
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

If P(A and B)=0 then two events are mutually exclusive.


STEP: How do we decide if events are mutually exclusive?
[−1 point ⇒ 3 / 4 points left]

Mutually exclusive events can never occur at the same time. For example, "Nneka is in the Western Cape" and "Nneka is in Gauteng" are mutually exclusive events, because Nneka cannot be in two places at once.

NOTE: Two events are mutually exclusive if the probability of both occuring at the same time is zero: P(A and B)=0.

STEP: Decide if the events are mutually exclusive
[−3 points ⇒ 0 / 4 points left]

For all events A and B:

P(A or B)=P(A)+P(B)P(A and B)

There are two ways to prove that events are mutually exclusive:

  1. Prove that P(A and B)=0.
  2. Prove that P(A or B)=P(A)+P(B).

We know that:

  • P(A)=0.4
  • P(B)=0.5, and
  • P(A or B)=0.8.

Method 1: Events A and B are mutually exclusive if P(A and B)=0

0.8=0.4+0.5P(A and B)
P(A and B)=0.10

Therefore the events are not mutually exclusive.

Method 2: Events A and B are mutually exclusive if P(A or B)=P(A)+P(B)

P(A)+P(B)=0.4+0.5=0.9P(A)+P(B)P(A or B)

Therefore the events are not mutually exclusive.

NOTE: Choose the method that you prefer, and use that one. You don't need to use both methods!

Of the given choices, Option 4 is correct.


Submit your answer as:

ID is: 4344 Seed is: 7696

Prove that two events are mutually exclusive

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

Given two events A and B:

  • P(A)=0.4
  • P(B)=0.3
  • P(A or B)=0.7

Are the events A and B mutually exclusive? Justify your answer with appropriate calculations.

INSTRUCTION: Choose the most correct option to answer this question.
Answer:
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

If P(A and B)=0 then two events are mutually exclusive.


STEP: How do we decide if events are mutually exclusive?
[−1 point ⇒ 3 / 4 points left]

Mutually exclusive events can never occur at the same time. For example, "Akinbode is in the Western Cape" and "Akinbode is in Gauteng" are mutually exclusive events, because Akinbode cannot be in two places at once.

NOTE: Two events are mutually exclusive if the probability of both occuring at the same time is zero: P(A and B)=0.

STEP: Decide if the events are mutually exclusive
[−3 points ⇒ 0 / 4 points left]

For all events A and B:

P(A or B)=P(A)+P(B)P(A and B)

There are two ways to prove that events are mutually exclusive:

  1. Prove that P(A and B)=0.
  2. Prove that P(A or B)=P(A)+P(B).

We know that:

  • P(A)=0.4
  • P(B)=0.3, and
  • P(A or B)=0.7.

Method 1: Events A and B are mutually exclusive if P(A and B)=0

0.7=0.4+0.3P(A and B)
P(A and B)=0

Therefore the events are mutually exclusive.

Method 2: Events A and B are mutually exclusive if P(A or B)=P(A)+P(B)

P(A)+P(B)=0.4+0.3=0.7P(A)+P(B)=P(A or B)

Therefore the events are mutually exclusive.

NOTE: Choose the method that you prefer, and use that one. You don't need to use both methods!

Of the given choices, Option 2 is correct.


Submit your answer as:

ID is: 4344 Seed is: 8513

Prove that two events are mutually exclusive

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

Given two events A and B:

  • P(A)=0.3
  • P(B)=0.5
  • P(A or B)=0.8

Are the events A and B mutually exclusive? Justify your answer with appropriate calculations.

INSTRUCTION: Choose the most correct option to answer this question.
Answer:
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

If P(A and B)=0 then two events are mutually exclusive.


STEP: How do we decide if events are mutually exclusive?
[−1 point ⇒ 3 / 4 points left]

Mutually exclusive events can never occur at the same time. For example, "Bukelwa is in the Western Cape" and "Bukelwa is in Gauteng" are mutually exclusive events, because Bukelwa cannot be in two places at once.

NOTE: Two events are mutually exclusive if the probability of both occuring at the same time is zero: P(A and B)=0.

STEP: Decide if the events are mutually exclusive
[−3 points ⇒ 0 / 4 points left]

For all events A and B:

P(A or B)=P(A)+P(B)P(A and B)

There are two ways to prove that events are mutually exclusive:

  1. Prove that P(A and B)=0.
  2. Prove that P(A or B)=P(A)+P(B).

We know that:

  • P(A)=0.3
  • P(B)=0.5, and
  • P(A or B)=0.8.

Method 1: Events A and B are mutually exclusive if P(A and B)=0

0.8=0.3+0.5P(A and B)
P(A and B)=0

Therefore the events are mutually exclusive.

Method 2: Events A and B are mutually exclusive if P(A or B)=P(A)+P(B)

P(A)+P(B)=0.3+0.5=0.8P(A)+P(B)=P(A or B)

Therefore the events are mutually exclusive.

NOTE: Choose the method that you prefer, and use that one. You don't need to use both methods!

Of the given choices, Option 1 is correct.


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ID is: 3140 Seed is: 150

Mutually exclusive events: phone numbers

In Pretoria phone numbers start with 012. After that there are seven more digits:

012-XXX-XXXX

With these seven digits, there are many phone numbers possible in Pretoria. Here are two statements about one of these seven-digit phone numbers:

  • The first digit in the phone number is 3.
  • The fifth and the seventh digits are both 7.
Answer: Are the statements mutually exclusive?
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two things are mutually exclusive if they cannot both be true at the same time. Can both statements be true about a single phone number?


STEP: Compare the statements to see if they can both be true
[−1 point ⇒ 0 / 1 points left]

The words 'mutually exclusive' mean that two things cannot both be true at the same time: if one of them is true the other must be false. In other words, there is a conflict between the statements. For example: 'The ANC has the majority in Parliament' and 'The IFP has the majority in Parliament.' If one of these statements is true, the other must be false. They are mutually exclusive statements because there is a conflict.

To decide if the statements are mutually exclusive, we must imagine one phone number which is seven digits long and ask ourselves: if one of the statements is true about that phone number, does that make the other statement false?

These statements are not mutually exclusive. We can even write down an example which agrees with both statements: 338-6737. If it is possible to write one phone number which agrees with both statements, they cannot be mutually exclusive.

The correct choice from the list is: No.


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ID is: 3140 Seed is: 4390

Mutually exclusive events: phone numbers

In Nelspruit phone numbers start with 013. After that there are seven more digits:

013-XXX-XXXX

With these seven digits, there are many phone numbers possible in Nelspruit. Here are two statements about one of these seven-digit phone numbers:

  • The phone number has the digit 7 six times.
  • The seventh digit in the phone number is 4.
Answer: Are the statements mutually exclusive?
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two things are mutually exclusive if they cannot both be true at the same time. Can both statements be true about a single phone number?


STEP: Compare the statements to see if they can both be true
[−1 point ⇒ 0 / 1 points left]

The words 'mutually exclusive' mean that two things cannot both be true at the same time: if one of them is true the other must be false. In other words, there is a conflict between the statements. For example: 'The ANC has the majority in Parliament' and 'The DA has the majority in Parliament.' If one of these statements is true, the other must be false. They are mutually exclusive statements because there is a conflict.

To decide if the statements are mutually exclusive, we must imagine one phone number which is seven digits long and ask ourselves: if one of the statements is true about that phone number, does that make the other statement false?

These statements are not mutually exclusive. We can even write down an example which agrees with both statements: 777-7774. If it is possible to write one phone number which agrees with both statements, they cannot be mutually exclusive.

The correct choice from the list is: No.


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ID is: 3140 Seed is: 1699

Mutually exclusive events: phone numbers

In Pretoria phone numbers start with 012. After that there are seven more digits:

012-XXX-XXXX

With these seven digits, there are many phone numbers possible in Pretoria. Here are two statements about one of these seven-digit phone numbers:

  • The phone number includes the digit 8 only once.
  • The phone number does not include the digit 8.
Answer: Are the statements mutually exclusive?
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two things are mutually exclusive if they cannot both be true at the same time. Can both statements be true about a single phone number?


STEP: Compare the statements to see if they can both be true
[−1 point ⇒ 0 / 1 points left]

The words 'mutually exclusive' mean that two things cannot both be true at the same time: if one of them is true the other must be false. In other words, there is a conflict between the statements. For example: 'The ANC has the majority in Parliament' and 'The IFP has the majority in Parliament.' If one of these statements is true, the other must be false. They are mutually exclusive statements because there is a conflict.

To decide if the statements are mutually exclusive, we must imagine one phone number which is seven digits long and ask ourselves: if one of the statements is true about that phone number, does that make the other statement false?

These statements are mutually exclusive: they conflict with each other. No phone number can include a 8 and at the same time not include a 8. (If one of the statements is true, the other must be false.)

The correct choice from the list is: Yes.


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ID is: 3136 Seed is: 3936

Mutually exclusive events

Answer the two questions below about mutually exclusive events.

  1. What does it mean if two events are 'mutually exclusive?'

    Answer: Mutually exclusive means: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Mutually exclusive statements conflict with each other. Here is an example of mutually exclusive statements; use them to choose the correct answer.

    1. My name starts with the letter A.
    2. My name starts with the letter K.

    STEP: Identify the statement which means the same thing as "mutually exclusive"
    [−1 point ⇒ 0 / 1 points left]

    The words "mutually exclusive" refer to two things that exclude each other. They cannot both be true at the same time. For example: "The ANC has the majority in Parliament" and "The IFP has the majority in Parliament." If one of these statements is true, the other must be false. They are mutually exclusive statements.

    The choice from the list which has the same meaning as this is: there is no chance that the events can happen together.


    Submit your answer as:
  2. Think about a person going into a shop where he can buy airtime and other basic things like milk, juice and chips. The following statements might describe what happens in the shop:

    • The person buys a single airtime voucher.
    • The person buys airtime from Virgin Mobile and Cell C.
    Answer: Are the statements mutually exclusive?
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Ask yourself this: can both of the statements be true when the person is in the shop? If the answer is no, then the statements are mutually exclusive.


    STEP: Compare the statements to see if they can both be true
    [−1 point ⇒ 0 / 1 points left]

    As stated above, two things are mutually exclusive if they cannot both be true at the same time. To decide if the statements are mutually exclusive, we must imagine that the person is in the shop and ask ourselves: if one of the statements is true, does that mean the other statement must be false? If so, the statements are mutually exclusive.

    These statements are mutually exclusive because they exclude each other. If the person only bought one airtime voucher, it cannot also be true that they bought airtime from 2 different service providers.

    The correct choice from the list is: Yes.


    Submit your answer as:

ID is: 3136 Seed is: 274

Mutually exclusive events

Answer the two questions below about mutually exclusive events.

  1. What does it mean if two events are 'mutually exclusive?'

    Answer: Mutually exclusive means: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Mutually exclusive statements conflict with each other. Here is an example of mutually exclusive statements; use them to choose the correct answer.

    1. My name starts with the letter A.
    2. My name starts with the letter K.

    STEP: Identify the statement which means the same thing as "mutually exclusive"
    [−1 point ⇒ 0 / 1 points left]

    The words "mutually exclusive" refer to two things that exclude each other. They cannot both be true at the same time. For example: "The ANC has the majority in Parliament" and "The EFF has the majority in Parliament." If one of these statements is true, the other must be false. They are mutually exclusive statements.

    The choice from the list which has the same meaning as this is: that two things cannot occur at the same time.


    Submit your answer as:
  2. Think about a person going into a shop where she can buy airtime and other basic things like milk, juice and chips. The following statements might describe what happens in the shop:

    • The person buys a single airtime voucher.
    • The person buys airtime from Vodacom and Telkom Mobile.
    Answer: Are the statements mutually exclusive?
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Ask yourself this: can both of the statements be true when the person is in the shop? If the answer is no, then the statements are mutually exclusive.


    STEP: Compare the statements to see if they can both be true
    [−1 point ⇒ 0 / 1 points left]

    As stated above, two things are mutually exclusive if they cannot both be true at the same time. To decide if the statements are mutually exclusive, we must imagine that the person is in the shop and ask ourselves: if one of the statements is true, does that mean the other statement must be false? If so, the statements are mutually exclusive.

    These statements are mutually exclusive because they exclude each other. If the person only bought one airtime voucher, it cannot also be true that they bought airtime from 2 different service providers.

    The correct choice from the list is: Yes.


    Submit your answer as:

ID is: 3136 Seed is: 5813

Mutually exclusive events

Answer the two questions below about mutually exclusive events.

  1. What does it mean if two events are 'mutually exclusive?'

    Answer: Mutually exclusive means: .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Mutually exclusive statements conflict with each other. Here is an example of mutually exclusive statements; use them to choose the correct answer.

    1. My name starts with the letter A.
    2. My name starts with the letter K.

    STEP: Identify the statement which means the same thing as "mutually exclusive"
    [−1 point ⇒ 0 / 1 points left]

    The words "mutually exclusive" refer to two things that exclude each other. They cannot both be true at the same time. For example: "The ANC has the majority in Parliament" and "The EFF has the majority in Parliament." If one of these statements is true, the other must be false. They are mutually exclusive statements.

    The choice from the list which has the same meaning as this is: that two things cannot occur at the same time.


    Submit your answer as:
  2. Think about a bus moving along toward a bus stop. The following statements describe what might happen when the bus comes to the bus stop:

    • The bus does not stop.
    • 4 people get off the bus.
    Answer: Are the statements mutually exclusive?
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Ask yourself this: can both of the statements be true when the bus comes to a stop? If the answer is no, then the statements are mutually exclusive.


    STEP: Compare the statements to see if they can both be true
    [−1 point ⇒ 0 / 1 points left]

    As stated above, two things are mutually exclusive if they cannot both be true at the same time. To decide if the statements are mutually exclusive, we must imagine that the bus comes to a stop and ask ourselves: if one of the statements is true, does that mean the other statement must be false? If so, the statements are mutually exclusive.

    These statements are mutually exclusive because they exclude each other. If the bus does not stop, it cannot be true that 4 people got off the bus.

    The correct choice from the list is: Yes.


    Submit your answer as:

4. Complementary events


ID is: 3141 Seed is: 3972

Complementary sets: basic facts

Consider a set A and a sample space S.

  1. For set A there is a complementary set, which has the name A. Choose the best option below about A.

    Answer:

    The set A ...

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Complementary sets do not share any elements.


    STEP: Use the definition of complementary sets to choose the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question describes a situation with three sets: the sample space S, another set A, and the complement of set A. A complementary set contains all of the elements in the sample space which are not in the original set. For example, if

    S={2;3;4;8}

    and

    A={3;4}

    then the complement of A is

    A={2;8}

    Two things are always true about complementary sets.

    1. The complementary set will never include any numbers contained in the original set.
    2. The complementary set will always include all of the elements of the sample space which are not in the original set.

    The correct choice from the list is that the set A includes all of the elements which are not in A.


    Submit your answer as:
  2. Which of the choices in the list is correct for the set (AA)? (Remember that S is the sample space.)

    Answer:

    AA=

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    This question asks about the union of sets A with A. Any set and its complement are like two sides of a coin: they are separate, but together they make the entire coin.


    STEP: Think about the union of the two sets
    [−1 point ⇒ 0 / 1 points left]

    Let's continue to use the example where S={2;3;4;8}.

    We know that A is the complement of A, so this means A has all the elements not in A.

    If A={3;4} then A={2;8}.

    We are asked about (AA). This is sometimes written as (A or A).

    (AA) means the elements that are in A or in A.

    AA={2;3;4;8}=S

    The correct answer is: S (the sample space).


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ID is: 3141 Seed is: 7195

Complementary sets: basic facts

Consider a set A and a sample space S.

  1. For set A there is a complementary set, which has the name A. Choose the best option below about A.

    Answer:

    The set A ...

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Complementary sets do not share any elements.


    STEP: Use the definition of complementary sets to choose the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question describes a situation with three sets: the sample space S, another set A, and the complement of set A. A complementary set contains all of the elements in the sample space which are not in the original set. For example, if

    S={2;3;6;7}

    and

    A={2;7}

    then the complement of A is

    A={3;6}

    Two things are always true about complementary sets.

    1. The complementary set will never include any numbers contained in the original set.
    2. The complementary set will always include all of the elements of the sample space which are not in the original set.

    The correct choice from the list is that the set A never includes the elements of A.


    Submit your answer as:
  2. Which of the choices in the list is correct for the set (AA)? (Remember that S is the sample space.)

    Answer:

    AA=

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    This question asks about the intersection of sets A with A. Any set and its complement are like two sides of a coin: they are separate, but together they make the entire coin.


    STEP: Think about the intersection of the two sets
    [−1 point ⇒ 0 / 1 points left]

    Let's continue to use the example where S={2;3;6;7}.

    We know that A is the complement of A, so this means A has all the elements not in A.

    If A={2;7} then A={3;6}.

    We are asked about (AA). This is sometimes written as (A and A).

    (AA) means the elements that are in A and in A.

    AA={}=

    The correct answer is: (the empty set).


    Submit your answer as:

ID is: 3141 Seed is: 3629

Complementary sets: basic facts

Consider a set A and a sample space S.

  1. For set A there is a complementary set, which has the name A. Choose the best option below about A.

    Answer:

    The set A ...

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Complementary sets do not share any elements.


    STEP: Use the definition of complementary sets to choose the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question describes a situation with three sets: the sample space S, another set A, and the complement of set A. A complementary set contains all of the elements in the sample space which are not in the original set. For example, if

    S={1;2;4;6;9;10}

    and

    A={1;2;10}

    then the complement of A is

    A={4;6;9}

    Two things are always true about complementary sets.

    1. The complementary set will never include any numbers contained in the original set.
    2. The complementary set will always include all of the elements of the sample space which are not in the original set.

    The correct choice from the list is that the set A never includes the elements of A.


    Submit your answer as:
  2. Which of the choices in the list is correct for the set (AA)? (Remember that S is the sample space.)

    Answer:

    AA=

    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    This question asks about the intersection of sets A with A. Any set and its complement are like two sides of a coin: they are separate, but together they make the entire coin.


    STEP: Think about the intersection of the two sets
    [−1 point ⇒ 0 / 1 points left]

    Let's continue to use the example where S={1;2;4;6;9;10}.

    We know that A is the complement of A, so this means A has all the elements not in A.

    If A={1;2;10} then A={4;6;9}.

    We are asked about (AA). This is sometimes written as (A and A).

    (AA) means the elements that are in A and in A.

    AA={}=

    The correct answer is: (the empty set).


    Submit your answer as:

ID is: 3134 Seed is: 9622

Complementary sets: identifying a complementary set

Consider this sample space:

S={1;2;3;7;8}

Here is a set in the sample space:

B={1;2;7}

Identify set B. Choose your answer from the list below.

Answer: B=
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The symbol B refers to the complement of set B. The complement of B must include every element in S which is not in B.


STEP: List the values which are in S but not in B
[−1 point ⇒ 0 / 1 points left]

In this question we need to identify the correct set for B, which is the complement of set B. A complementary set includes all of the elements which are not included in the original set. Another way to think about it is like this: a set together with its complement will always contain all of the elements in the sample space.

In this case, we have the sample space:

S={1;2;3;7;8}

and the set:

B={1;2;7}

To find B, take the elements in B out of S. Whatever elements remain belong in B:

B={3;8}

In this case, there are 2 elements which remain.

The correct choice from the list is: {3;8}.


Submit your answer as:

ID is: 3134 Seed is: 9267

Complementary sets: identifying a complementary set

Consider this sample space:

S={1;2;4;7;8;9}

Here is a set in the sample space:

G={2;8}

Identify set G. Choose your answer from the list below.

Answer: G=
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The symbol G refers to the complement of set G. The complement of G must include every element in S which is not in G.


STEP: List the values which are in S but not in G
[−1 point ⇒ 0 / 1 points left]

In this question we need to identify the correct set for G, which is the complement of set G. A complementary set includes all of the elements which are not included in the original set. Another way to think about it is like this: a set together with its complement will always contain all of the elements in the sample space.

In this case, we have the sample space:

S={1;2;4;7;8;9}

and the set:

G={2;8}

To find G, take the elements in G out of S. Whatever elements remain belong in G:

G={1;4;7;9}

In this case, there are 4 elements which remain.

The correct choice from the list is: {1;4;7;9}.


Submit your answer as:

ID is: 3134 Seed is: 3670

Complementary sets: identifying a complementary set

Consider this sample space:

S={1;6;7;8;9;10}

Here is a set in the sample space:

G={}

Identify set G. Choose your answer from the list below.

Answer: G=
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

The symbol G refers to the complement of set G. The complement of G must include every element in S which is not in G.


STEP: List the values which are in S but not in G
[−1 point ⇒ 0 / 1 points left]

In this question we need to identify the correct set for G, which is the complement of set G. A complementary set includes all of the elements which are not included in the original set. Another way to think about it is like this: a set together with its complement will always contain all of the elements in the sample space.

In this case, we have the sample space:

S={1;6;7;8;9;10}

and the set:

G={}

To find G, take the elements in G out of S. Whatever elements remain belong in G:

G={1;6;7;8;9;10}

In this case, there are 6 elements which remain. It turns out that G contains all of the elements in S. This is true because set G has no numbers in it at all.

The correct choice from the list is: {1;6;7;8;9;10}.


Submit your answer as:

ID is: 3165 Seed is: 9759

Venn diagram: complementary sets

The Venn diagram below shows sets A, B, and C.

Identify set (A and B and C) from the drop down list:

Answer: The correct set is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Complementary events are two outcomes of an event that are the only two possible outcomes. If the one outcome is the A and B and C, what is the only other possible outcome?


STEP: Which region or regions of the Venn diagram hold the elements A and B and C? What is left over?
[−1 point ⇒ 0 / 1 points left]

The question asked for: (A and B and C). In other words the complementary set of A and B and C. So the first step is to identify the A and B and C region on the diagram. The yellow shaded area below represents the region A and B and C:

Represented as a set:

A and B and C={K}

A complementary set includes all of the elements which are not included in the set A and B and C. If we take out all the elements from the shaded area above, then we are left with the only other elements, in the complementary set of A and B and C.

(A and B and C)={H;V;b;e;f;g;h;j;n;t}

If you look at the diagram again this is all the elements which are in the green shaded area.

Therefore the correct set for (A and B and C)={H;V;b;e;f;g;h;j;n;t}.


Submit your answer as:

ID is: 3165 Seed is: 9945

Venn diagram: complementary sets

The Venn diagram below shows sets A, B, and C.

Identify set (A or B) from the drop down list:

Answer: The correct set is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Complementary events are two outcomes of an event that are the only two possible outcomes. If the one outcome is the A or B, what is the only other possible outcome?


STEP: Which region or regions of the Venn diagram hold the elements A or B? What is left over?
[−1 point ⇒ 0 / 1 points left]

The question asked for: (A or B). In other words the complementary set of A or B. So the first step is to identify the A or B region on the diagram. The yellow shaded area below represents the region A or B:

Represented as a set:

A or B={12;16;2;26;35;38;46;50;54;8}

A complementary set includes all of the elements which are not included in the set A or B. If we take out all the elements from the shaded area above, then we are left with the only other elements, in the complementary set of A or B.

(A or B)={18;19;20;24}

If you look at the diagram again this is all the elements which are in the green shaded area.

Therefore the correct set for (A or B)={18;19;20;24}.


Submit your answer as:

ID is: 3165 Seed is: 262

Venn diagram: complementary sets

The Venn diagram below shows sets A, B, and C.

Identify set (A or B or C) from the drop down list:

Answer: The correct set is .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Complementary events are two outcomes of an event that are the only two possible outcomes. If the one outcome is the A or B or C, what is the only other possible outcome?


STEP: Which region or regions of the Venn diagram hold the elements A or B or C? What is left over?
[−1 point ⇒ 0 / 1 points left]

The question asked for: (A or B or C). In other words the complementary set of A or B or C. So the first step is to identify the A or B or C region on the diagram. The yellow shaded area below represents the region A or B or C:

Represented as a set:

A or B or C={E;F;N;S;Y;a;h;j;k;m;q;r;z}

A complementary set includes all of the elements which are not included in the set A or B or C. If we take out all the elements from the shaded area above, then we are left with the only other elements, in the complementary set of A or B or C.

(A or B or C)={}

There are no elements in this set. The set A or B or C holds all the available elements in the Venn diagram so there are no other elements left for (A or B or C).

Therefore the correct set for (A or B or C)={}.


Submit your answer as:

ID is: 1192 Seed is: 4385

Working with complementary events

A group of students are given the following Venn diagram:

The sample space can be described as {n:n ϵ Z, 1n15}.

They want to find the complement of (A and B). (This complement is also known as (A and B).) They get stuck, and you offer to help them find it.

Which of the choices below shows the correct event set of (A and B)?

Answer: (A and B)=
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by figuring out the elements of set (A and B). Then find the complement.


STEP: Identify the elements in the complementary set
[−1 point ⇒ 0 / 1 points left]

The event set (A and B) can be shaded as follows:

The complement is every other part of the Venn diagram. So event set (A and B) can be shaded as follows:

The complement of (A and B) is (A and B)= {1;2;3;5;6;7;9;10;11;12;13;14;15}.


Submit your answer as:

ID is: 1192 Seed is: 3883

Working with complementary events

A group of students are given the following Venn diagram:

The sample space can be described as {n:n ϵ Z, 1n15}.

They want to find the complement of (A and B). (This complement is also known as (A and B).) They get stuck, and you offer to help them find it.

Which of the choices below shows the correct event set of (A and B)?

Answer: (A and B)=
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by figuring out the elements of set (A and B). Then find the complement.


STEP: Identify the elements in the complementary set
[−1 point ⇒ 0 / 1 points left]

The event set (A and B) can be shaded as follows:

The complement is every other part of the Venn diagram. So event set (A and B) can be shaded as follows:

The complement of (A and B) is (A and B)= {1;3;4;8;9;10;11;12;13;14;15}.


Submit your answer as:

ID is: 1192 Seed is: 7519

Working with complementary events

A group of students are given the following Venn diagram:

The sample space can be described as {n:n ϵ Z, 1n15}.

They want to find the complement of B. (This complement is also known as B.) They get stuck, and you offer to help them find it.

Which of the choices below shows the correct event set of B?

Answer: B=
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by figuring out the elements of set B. Then find the complement.


STEP: Identify the elements in the complementary set
[−1 point ⇒ 0 / 1 points left]

The event set B can be shaded as follows:

The complement is every other part of the Venn diagram. So event set B can be shaded as follows:

The complement of B is B= {2;3;4;5;6;7;8;11;13;15}.


Submit your answer as:

ID is: 3139 Seed is: 6844

Complementary sets: working with set notation

Consider the following sample space:

S={s:sZ,4s8}
  1. How many elements are in S?

    Answer: The number of elements in S is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    What is the smallest number in S? What is the largest number in S? If you know that, you can figure out how many elements are in the set.


    STEP: Write out the elements in the set to determine how many there are
    [−1 point ⇒ 0 / 1 points left]

    We have the set

    S={s:sZ,4s8}

    and we need to figure out how many elements are in it. This means we need to find out how many numbers are in the set.

    We need to interpret the set given to find out what numbers belong in the set. First, notice that the elements of S are integers because "sZ". This alone means that the values come from:

    Z={3;2;1;0;1;2;3}

    But set S also says that 4s8. This means that set S only includes numbers in Z from 4 up to and including 8. So the sample space, set S, includes these numbers:

    S={4;3;2;1;0;1;2;3;4;5;6;7;8}

    Now count the numbers in the set to get the answer. We can see clearly that there are 13 numbers in set S.

    There are 13 elements in S.


    Submit your answer as:
  2. Within the sample space of set S, suppose there is a set:

    B={4;1;5;6;8}

    Give any element in B. If there are no elements in B write none in the answer box.

    INSTRUCTION: You should enter a single number from B or the word none.
    Answer: One of the elements in B is: .
    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The symbol B refers to the complement of set B with respect to the sample space in Question 1.


    STEP: Determine the complement of set B and pick one of its values for the answer
    [−1 point ⇒ 0 / 1 points left]

    The question shows us set B, which is a subset of the sample space S. We need to identify an element which is in B, which is the complement of B. The complement of a set includes all of the values from the sample space which are not in that set.

    We can answer this question by comparing the list in Question 1 to set B. The complement of B includes all of the numbers in S which are not in B.

    S={4;3;2;1;0;1;2;3;4;5;6;7;8}andB={4;1;5;6;8}

    Comparing these sets, we can see that the complement of B includes these 8 numbers:

    B={3;2;0;1;2;3;4;7}

    Therefore, any one of the following numbers is an acceptable answer: -3 ; -2 ; 0 ; 1 ; 2 ; 3 ; 4 ; 7.


    Submit your answer as:

ID is: 3139 Seed is: 5672

Complementary sets: working with set notation

Consider the following sample space:

S={s:sN0,s7}
  1. How many elements are in S?

    Answer: The number of elements in S is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    What is the smallest number in S? What is the largest number in S? If you know that, you can figure out how many elements are in the set.


    STEP: Write out the elements in the set to determine how many there are
    [−1 point ⇒ 0 / 1 points left]

    We have the set

    S={s:sN0,s7}

    and we need to figure out how many elements are in it. This means we need to find out how many numbers are in the set.

    We need to interpret the set given to find out what numbers belong in the set. First, notice that the elements of S are whole (counting) numbers because "sN0". This alone means that the values come from:

    N0={0;1;2;3;4}

    But set S also says that s7. This means that set S only includes numbers in N0 up to and including 7. So the sample space, set S, includes these numbers:

    S={0;1;2;3;4;5;6;7}

    Now count the numbers in the set to get the answer. We can see clearly that there are 8 numbers in set S.

    There are 8 elements in S.


    Submit your answer as:
  2. Within the sample space of set S, suppose there is a set:

    B={0;1;3;5}

    Give any element in B. If there are no elements in B write none in the answer box.

    INSTRUCTION: You should enter a single number from B or the word none.
    Answer: One of the elements in B is: .
    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The symbol B refers to the complement of set B with respect to the sample space in Question 1.


    STEP: Determine the complement of set B and pick one of its values for the answer
    [−1 point ⇒ 0 / 1 points left]

    The question shows us set B, which is a subset of the sample space S. We need to identify an element which is in B, which is the complement of B. The complement of a set includes all of the values from the sample space which are not in that set.

    We can answer this question by comparing the list in Question 1 to set B. The complement of B includes all of the numbers in S which are not in B.

    S={0;1;2;3;4;5;6;7}andB={0;1;3;5}

    Comparing these sets, we can see that the complement of B includes these 4 numbers:

    B={2;4;6;7}

    Therefore, any one of the following numbers is an acceptable answer: 2 ; 4 ; 6 ; 7.


    Submit your answer as:

ID is: 3139 Seed is: 9913

Complementary sets: working with set notation

Consider the following sample space:

S={x:xZ,2<x<8}
  1. How many elements are in S?

    Answer: The number of elements in S is .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    What is the smallest number in S? What is the largest number in S? If you know that, you can figure out how many elements are in the set.


    STEP: Write out the elements in the set to determine how many there are
    [−1 point ⇒ 0 / 1 points left]

    We have the set

    S={x:xZ,2<x<8}

    and we need to figure out how many elements are in it. This means we need to find out how many numbers are in the set.

    We need to interpret the set given to find out what numbers belong in the set. First, notice that the elements of S are integers because "xZ". This alone means that the values come from:

    Z={3;2;1;0;1;2;3}

    But set S also says that 2<x<8. This means that set S only includes numbers in Z greater than 2 and less than 8. So the sample space, set S, includes these numbers:

    S={1;0;1;2;3;4;5;6;7}

    Now count the numbers in the set to get the answer. We can see clearly that there are 9 numbers in set S.

    There are 9 elements in S.


    Submit your answer as:
  2. Within the sample space of set S, suppose there is a set:

    X={1;1;2;3;5;6;7}

    Give any element in X. If there are no elements in X write none in the answer box.

    INSTRUCTION: You should enter a single number from X or the word none.
    Answer: One of the elements in X is: .
    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The symbol X refers to the complement of set X with respect to the sample space in Question 1.


    STEP: Determine the complement of set X and pick one of its values for the answer
    [−1 point ⇒ 0 / 1 points left]

    The question shows us set X, which is a subset of the sample space S. We need to identify an element which is in X, which is the complement of X. The complement of a set includes all of the values from the sample space which are not in that set.

    We can answer this question by comparing the list in Question 1 to set X. The complement of X includes all of the numbers in S which are not in X.

    S={1;0;1;2;3;4;5;6;7}andX={1;1;2;3;5;6;7}

    Comparing these sets, we can see that the complement of X includes these 2 numbers:

    X={0;4}

    Therefore, any one of the following numbers is an acceptable answer: 0 ; 4.


    Submit your answer as:

5. Independent events


ID is: 4342 Seed is: 8245

Contingency tables: proving whether events are independent

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

The table below shows data on the monthly income of employed people in two residential areas. Representative samples were used in the collection of the data.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 1,867 800 2,667
3,200x<25,600 1,444 438 1,882
x25,600 84 218 302
Total 3,395 1,456 4,851

Is the area in which a person lives independent, or not independent, of earning a monthly income of less than N=3,200?

INSTRUCTION: Round your answers to two decimal places, and compare these rounded values to answer the question.
Answer:

Choose the correct expressions to prove whether the events are independent or not, and calculate their values. (You must choose appropriate events for A and B.)

  • =
  • =

Therefore,

So, earning a monthly income of less than N=3,200 is of the area in which a person lives.

one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Events A and B are independent if P(A and B)=P(A)×P(B).


STEP: What do we need to calculate to decide if events are independent or not?
[−1 point ⇒ 4 / 5 points left]
Events A and B are independent if P(A and B)=P(A)×P(B).

Let the events be:

  • A: People who earn less than N=3,200
  • B: People who live in Area 1

So the event A and B is: people who live in Area 1 and earn less than N=3,200. To calculate the probability of each event, we divide by the grand total.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 1,867 800 2,667
3,200x<25,600 1,444 438 1,882
x25,600 84 218 302
Total 3,395 1,456 4,851
TIP: We could also let Event B be "People who live in Area 2". The calculations would be different, but the conclusion would be the same.

STEP: Calculate the probablity that a person earns less than N=3,200 and lives in Area 1
[−1 point ⇒ 3 / 5 points left]
P(A and B)=1,8674,851=0.38486...0.38

STEP: Calculate P(lives in Area 1)×P(earns less than N=3,200)
[−2 points ⇒ 1 / 5 points left]
P(A)×P(B)=3,3954,851×2,6674,851=0.38476...0.38

STEP: Compare the two quantities when rounded to two decimal places
[−1 point ⇒ 0 / 5 points left]
P(A and B)=P(A)×P(B)

Therefore the events are independent.

NOTE: If you look at further decimal places, you might notice that the values of P(A and B) and P(A)×P(B) are not identical. However, they are very close to each other (equivalent to two decimal places), which is enough to conclude that the events are independent.

Submit your answer as: andandandand
and

ID is: 4342 Seed is: 7459

Contingency tables: proving whether events are independent

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

The table below shows data on the monthly income of employed people in two residential areas. Representative samples were used in the collection of the data.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 1,433 1,347 2,780
3,200x<25,600 717 1,079 1,796
x25,600 55 269 324
Total 2,205 2,695 4,900

Is earning a montly income of less than N=3,200 per month independent, or not independent, of the area in which a person lives?

INSTRUCTION: Round your answers to two decimal places, and compare these rounded values to answer the question.
Answer:

Choose the correct expressions to prove whether the events are independent or not, and calculate their values. (You must choose appropriate events for A and B.)

  • =
  • =

Therefore,

So, earning a monthly income of less than N=3,200 is of the area in which a person lives.

one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Events A and B are independent if P(A and B)=P(A)×P(B).


STEP: What do we need to calculate to decide if events are independent or not?
[−1 point ⇒ 4 / 5 points left]
Events A and B are independent if P(A and B)=P(A)×P(B).

Let the events be:

  • A: People who earn less than N=3,200
  • B: People who live in Area 1

So the event A and B is: people who live in Area 1 and earn less than N=3,200. To calculate the probability of each event, we divide by the grand total.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 1,433 1,347 2,780
3,200x<25,600 717 1,079 1,796
x25,600 55 269 324
Total 2,205 2,695 4,900
TIP: We could also let Event B be "People who live in Area 2". The calculations would be different, but the conclusion would be the same.

STEP: Calculate the probablity that a person earns less than N=3,200 and lives in Area 1
[−1 point ⇒ 3 / 5 points left]
P(A and B)=1,4334,900=0.29244...0.29

STEP: Calculate P(lives in Area 1)×P(earns less than N=3,200)
[−2 points ⇒ 1 / 5 points left]
P(A)×P(B)=2,2054,900×2,7804,900=0.25530...0.26

STEP: Compare the two quantities when rounded to two decimal places
[−1 point ⇒ 0 / 5 points left]
P(A and B)P(A)×P(B)

Therefore the events are not independent.


Submit your answer as: andandandand
and

ID is: 4342 Seed is: 1086

Contingency tables: proving whether events are independent

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

The table below shows data on the monthly income of employed people in two residential areas. Representative samples were used in the collection of the data.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 427 190 617
3,200x<25,600 856 666 1,522
x25,600 1,568 1,046 2,614
Total 2,851 1,902 4,753

Is earning a montly income of at least N=25,600 per month independent, or not independent, of the area in which a person lives?

INSTRUCTION: Round your answers to two decimal places, and compare these rounded values to answer the question.
Answer:

Choose the correct expressions to prove whether the events are independent or not, and calculate their values. (You must choose appropriate events for A and B.)

  • =
  • =

Therefore,

So, earning a monthly income of at least N=25,600 is of the area in which a person lives.

one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Events A and B are independent if P(A and B)=P(A)×P(B).


STEP: What do we need to calculate to decide if events are independent or not?
[−1 point ⇒ 4 / 5 points left]
Events A and B are independent if P(A and B)=P(A)×P(B).

Let the events be:

  • A: People who earn at least N=25,600
  • B: People who live in Area 1

So the event A and B is: people who live in Area 1 and earn at least N=25,600. To calculate the probability of each event, we divide by the grand total.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 427 190 617
3,200x<25,600 856 666 1,522
x25,600 1,568 1,046 2,614
Total 2,851 1,902 4,753
TIP: We could also let Event B be "People who live in Area 2". The calculations would be different, but the conclusion would be the same.

STEP: Calculate the probablity that a person earns at least N=25,600 and lives in Area 1
[−1 point ⇒ 3 / 5 points left]
P(A and B)=1,5684,753=0.32989...0.33

STEP: Calculate P(lives in Area 1)×P(earns at least N=25,600)
[−2 points ⇒ 1 / 5 points left]
P(A)×P(B)=2,8514,753×2,6144,753=0.32988...0.33

STEP: Compare the two quantities when rounded to two decimal places
[−1 point ⇒ 0 / 5 points left]
P(A and B)=P(A)×P(B)

Therefore the events are independent.

NOTE: If you look at further decimal places, you might notice that the values of P(A and B) and P(A)×P(B) are not identical. However, they are very close to each other (equivalent to two decimal places), which is enough to conclude that the events are independent.

Submit your answer as: andandandand
and

ID is: 3971 Seed is: 1132

Dependent and independent events

Adapted from DBE Nov 2015 Grade 12, P1, Q11.1
Maths formulas

For two events C and D, it is given that:

  • P(C)=0.6
  • P(D)=0.15
  • P(C and D)=0.483

Are the events, C and D, dependent or independent? Perform an appropriate calculation and construct a reason for your answer using the pairs of quantities in the table below:

A P(C or D) P(C)×P(D)
B P(C or D) P(C)+P(D)
C P(C and D) P(C)+P(D)
D P(C and D) P(C)×P(D)
Answer:

C and D are , because the quantities in Row are .

2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You may find it helpful to revise dependent and independent events in the Everything Maths textbook.


STEP: Determine whether C and D are independent
[−3 points ⇒ 0 / 3 points left]

Two events X and Y are independent if

P(X)×P(Y)=P(X and Y)

In our question:

P(C)×P(D)=0.6×0.15=0.09P(C and D)=0.483

The values are not equal, so the events C and D are dependent.


Submit your answer as: andand

ID is: 3971 Seed is: 3878

Dependent and independent events

Adapted from DBE Nov 2015 Grade 12, P1, Q11.1
Maths formulas

For two events M and N, it is given that:

  • P(M)=0.85
  • P(N)=0.3
  • P(M and N)=0.255

Are the events, M and N, dependent or independent? Perform an appropriate calculation and construct a reason for your answer using the pairs of quantities in the table below:

A P(M and N) P(M)×P(N)
B P(M or N) P(M)×P(N)
C P(M and N) P(M)+P(N)
D P(M or N) P(M)+P(N)
Answer:

M and N are , because the quantities in Row are .

2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You may find it helpful to revise dependent and independent events in the Everything Maths textbook.


STEP: Determine whether M and N are independent
[−3 points ⇒ 0 / 3 points left]

Two events X and Y are independent if

P(X)×P(Y)=P(X and Y)

In our question:

P(M)×P(N)=0.85×0.3=0.255P(M and N)=0.255

The values are equal, so the events M and N are independent.


Submit your answer as: andand

ID is: 3971 Seed is: 2275

Dependent and independent events

Adapted from DBE Nov 2015 Grade 12, P1, Q11.1
Maths formulas

For two events G and H, it is given that:

  • P(G)=0.22
  • P(H)=0.75
  • P(G and H)=0.681

Are the events, G and H, dependent or independent? Perform an appropriate calculation and construct a reason for your answer using the pairs of quantities in the table below:

A P(G and H) P(G)+P(H)
B P(G or H) P(G)×P(H)
C P(G or H) P(G)+P(H)
D P(G and H) P(G)×P(H)
Answer:

G and H are , because the quantities in Row are .

2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You may find it helpful to revise dependent and independent events in the Everything Maths textbook.


STEP: Determine whether G and H are independent
[−3 points ⇒ 0 / 3 points left]

Two events X and Y are independent if

P(X)×P(Y)=P(X and Y)

In our question:

P(G)×P(H)=0.22×0.75=0.165P(G and H)=0.681

The values are not equal, so the events G and H are dependent.


Submit your answer as: andand

6. Practical applications


ID is: 2303 Seed is: 9541

Two-way tables

There were customers in a restaurant that were asked whether they liked chicken or beef and whether they liked rice or pasta. The results of the survey are summarised in the following two-way frequency table.

Chicken Beef Totals
Rice 95 31 126
Pasta a 46 65
Totals 114 b c

Determine the values of the unknown quantities: a, b, and c.

Answer:
  1. a=
  2. b=
  3. c=
numeric
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

In a frequency table:

  • the totals of each row appear at the right;
  • the totals of each column appear at the bottom.
  • sum of the row totals equals the sum of the column totals. In this table, this grand total is c (seen in the lower right corner).

Use this information to find the values of the missing quantities.


STEP: Calculate the missing values: a, b, and c
[−3 points ⇒ 0 / 3 points left]

We have a two-way frequency table that came from asking the customers in a restaurant whether they liked chicken or beef and whether they liked rice or pasta.

Two-way frequency tables provide a visual representation of the relationships between grouped data. In a frequency table:

  • the totals of each row appear at the right;
  • the totals of each column appear at the bottom.
  • sum of the row totals equals the sum of the column totals. In this table, this grand total is c (seen in the lower right corner).

Using this information, we can calculate the missing values: a, b, and c. We do this by setting up equations for these quantities. We can set up the equations using row values or column values.

Calculate the missing value a:

a+46=65a=6546a=19

Calculate the missing value b:

31+46=b77=b

Calculate the missing value c:

126+65=c191=c

Therefore the completed table is:

Chicken Beef Totals
Rice 95 31 126
Pasta 19 46 65
Totals 114 77 191

The correct answers are:

  1. a=19
  2. b=77
  3. c=191

Submit your answer as: andand

ID is: 2303 Seed is: 8194

Two-way tables

There were customers in a restaurant that were asked whether they liked chicken or beef and whether they liked rice or pasta. The results of the survey are summarised in the following two-way frequency table.

Chicken Beef Totals
Rice 38 x 47
Pasta 27 8 y
Totals z 17 82

Determine the values of the unknown quantities: x, y, and z.

Answer:
  1. x=
  2. y=
  3. z=
numeric
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

In a frequency table:

  • the totals of each row appear at the right;
  • the totals of each column appear at the bottom.
  • sum of the row totals equals the sum of the column totals. In this table, this grand total is 82 (seen in the lower right corner).

Use this information to find the values of the missing quantities.


STEP: Calculate the missing values: x, y, and z
[−3 points ⇒ 0 / 3 points left]

We have a two-way frequency table that came from asking the customers in a restaurant whether they liked chicken or beef and whether they liked rice or pasta.

Two-way frequency tables provide a visual representation of the relationships between grouped data. In a frequency table:

  • the totals of each row appear at the right;
  • the totals of each column appear at the bottom.
  • sum of the row totals equals the sum of the column totals. In this table, this grand total is 82 (seen in the lower right corner).

Using this information, we can calculate the missing values: x, y, and z. We do this by setting up equations for these quantities. We can set up the equations using row values or column values.

Calculate the missing value x:

38+x=47x=4738x=9

Calculate the missing value y:

27+8=y35=y

Calculate the missing value z:

z+17=82z=8217z=65

Therefore the completed table is:

Chicken Beef Totals
Rice 38 9 47
Pasta 27 8 35
Totals 65 17 82

The correct answers are:

  1. x=9
  2. y=35
  3. z=65

Submit your answer as: andand

ID is: 2303 Seed is: 3193

Two-way tables

The number of Grade 11 and 12 students taking Accounting and History at Macibe High School are given in the following two-way frequency table.

Accounting History Totals
11 a b c
12 83 42 125
Totals 91 91 182

Determine the values of the unknown quantities: a, b, and c.

Answer:
  1. a=
  2. b=
  3. c=
numeric
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

In a frequency table:

  • the totals of each row appear at the right;
  • the totals of each column appear at the bottom.
  • sum of the row totals equals the sum of the column totals. In this table, this grand total is 182 (seen in the lower right corner).

Use this information to find the values of the missing quantities.


STEP: Calculate the missing values: a, b, and c
[−3 points ⇒ 0 / 3 points left]

We have a two-way frequency table that came from the records of the number of students taking Accounting and History at Macibe High School.

Two-way frequency tables provide a visual representation of the relationships between grouped data. In a frequency table:

  • the totals of each row appear at the right;
  • the totals of each column appear at the bottom.
  • sum of the row totals equals the sum of the column totals. In this table, this grand total is 182 (seen in the lower right corner).

Using this information, we can calculate the missing values: a, b, and c. We do this by setting up equations for these quantities. We can set up the equations using row values or column values.

Calculate the missing value a:

a+83=91a=9183a=8

Calculate the missing value b:

b+42=91b=9142b=49

Calculate the missing value c:

c+125=182c=182125c=57

Therefore the completed table is:

Accounting History Totals
11 8 49 57
12 83 42 125
Totals 91 91 182

The correct answers are:

  1. a=8
  2. b=49
  3. c=57

Submit your answer as: andand

ID is: 1644 Seed is: 4931

Tree diagrams with replacement

A bag contains 3 blue balls and 5 red balls. Habubakar picks a ball at random from the bag and replaces it back in the bag. He mixes the balls in the bag and then picks another ball at random from the bag.

The following tree diagram has been constructed using the information above.

Determine the values of the missing probabilities: X, Z, and V.

INSTRUCTION: Give you answer as a simplified fraction.
Answer:
  1. X=
  2. Z=
  3. V=
fraction
fraction
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question tells you that the first ball picked from the bag is replaced before the second ball is picked. This means that the total probability of each pair of branches is fixed.


STEP: Determine the values of X, Z, and V
[−3 points ⇒ 0 / 3 points left]

Tree diagrams are a way to visualise all the possible outcomes of an event or set of events. Each branch in the diagram represents a possible outcome.

To complete the given tree diagram, we need to calculate both the probability of picking a blue ball, and the probability of picking a red ball.

We calculate the probability of an event as follows:

P(outcome)=favourable outcomesNumber ofpossible outcomesTotal number of

The question tells us that the first ball picked from the bag is replaced before the second ball is picked. This means that the number of favourable outcomes, and the total number of possible outcomes, are both fixed, since we are always picking from the same set of balls.

The probability of picking a blue ball is:

P(blue)=blue ballsNumber ofballsTotal number of=38

The probability of picking a red ball is:

P(red)=red ballsNumber ofballsTotal number of=58

Using these results, we can fill in the missing probabilities on the tree diagram. The completed diagram is:

The values of X, Z, and V are:

  1. X=38
  2. Z=58
  3. V=58

Submit your answer as: andand

ID is: 1644 Seed is: 9195

Tree diagrams with replacement

A bag contains 7 blue balls and 6 white balls. Daniel picks a ball at random from the bag and replaces it back in the bag. He mixes the balls in the bag and then picks another ball at random from the bag.

The following tree diagram has been constructed using the information above.

Determine the values of the missing probabilities: L, M, and H.

INSTRUCTION: Give you answer as a simplified fraction.
Answer:
  1. L=
  2. M=
  3. H=
fraction
fraction
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question tells you that the first ball picked from the bag is replaced before the second ball is picked. This means that the total probability of each pair of branches is fixed.


STEP: Determine the values of L, M, and H
[−3 points ⇒ 0 / 3 points left]

Tree diagrams are a way to visualise all the possible outcomes of an event or set of events. Each branch in the diagram represents a possible outcome.

To complete the given tree diagram, we need to calculate both the probability of picking a blue ball, and the probability of picking a white ball.

We calculate the probability of an event as follows:

P(outcome)=favourable outcomesNumber ofpossible outcomesTotal number of

The question tells us that the first ball picked from the bag is replaced before the second ball is picked. This means that the number of favourable outcomes, and the total number of possible outcomes, are both fixed, since we are always picking from the same set of balls.

The probability of picking a blue ball is:

P(blue)=blue ballsNumber ofballsTotal number of=713

The probability of picking a white ball is:

P(white)=white ballsNumber ofballsTotal number of=613

Using these results, we can fill in the missing probabilities on the tree diagram. The completed diagram is:

The values of L, M, and H are:

  1. L=713
  2. M=713
  3. H=613

Submit your answer as: andand

ID is: 1644 Seed is: 3402

Tree diagrams with replacement

There are only blue marbles and yellow marbles in a bag. There are 2 blue marbles and 7 yellow marbles. Ayomide takes a marble from the bag at random. He puts the marble back in the bag. He then takes another random marble from the bag.

The following tree diagram has been constructed using the information above.

Determine the values of the missing probabilities: W, M, and K.

INSTRUCTION: Give you answer as a simplified fraction.
Answer:
  1. W=
  2. M=
  3. K=
fraction
fraction
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question tells you that the first marble picked from the bag is replaced before the second marble is picked. This means that the total probability of each pair of branches is fixed.


STEP: Determine the values of W, M, and K
[−3 points ⇒ 0 / 3 points left]

Tree diagrams are a way to visualise all the possible outcomes of an event or set of events. Each branch in the diagram represents a possible outcome.

To complete the given tree diagram, we need to calculate both the probability of picking a blue marble, and the probability of picking a yellow marble.

We calculate the probability of an event as follows:

P(outcome)=favourable outcomesNumber ofpossible outcomesTotal number of

The question tells us that the first marble picked from the bag is replaced before the second marble is picked. This means that the number of favourable outcomes, and the total number of possible outcomes, are both fixed, since we are always picking from the same set of marbles.

The probability of picking a blue marble is:

P(blue)=blue marblesNumber ofmarblesTotal number of=29

The probability of picking a yellow marble is:

P(yellow)=yellow marblesNumber ofmarblesTotal number of=79

Using these results, we can fill in the missing probabilities on the tree diagram. The completed diagram is:

The values of W, M, and K are:

  1. W=29
  2. M=29
  3. K=79

Submit your answer as: andand

ID is: 1648 Seed is: 609

Contingency tables

Learners in Grade 11 were interviewed about their favourite animal and the results of the survey were recorded in the following two-way table:

Dog Cat Total
Girls x y z
Boys 56 11 67
Total 104 100 204
  1. What is the value of x?

    Answer: x=
    numeric
    2 attempts remaining
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In a two-way table, the numbers in each column have to add up to the total for that column. In the same way, the numbers in each row have to add up to the total for that row. Looking at the first column:

    x+56=104x=48

    Submit your answer as:
  2. What is the value of y?

    Answer: y=
    numeric
    2 attempts remaining
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In a two-way table, the numbers in each column have to add up to the total for that column. In the same way, the numbers in each row have to add up to the total for that row. Looking at the second column:

    y+11=100y=89

    Submit your answer as:
  3. What is the value of z?

    Answer: z=
    numeric
    2 attempts remaining
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In a two-way table, the numbers in each column have to add up to the total for that column. In the same way, the numbers in each row have to add up to the total for that row. Looking at the third column:

    z+67=204z=137

    Submit your answer as:

ID is: 1648 Seed is: 1316

Contingency tables

Learners in Grade 11 were interviewed about their favourite cooldrink and the results of the survey were recorded in the following two-way table:

Coke Fanta Total
Girls 60 5 65
Boys 8 z 59
Total x y 124
  1. What is the value of x?

    Answer: x=
    numeric
    2 attempts remaining
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In a two-way table, the numbers in each column have to add up to the total for that column. In the same way, the numbers in each row have to add up to the total for that row. Looking at the first column:

    60+8=xx=68

    Submit your answer as:
  2. What is the value of y?

    Answer: y=
    numeric
    2 attempts remaining
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In a two-way table, the numbers in each column have to add up to the total for that column. In the same way, the numbers in each row have to add up to the total for that row. Looking at the third row:

    x+y=12468+y=124y=56

    Submit your answer as:
  3. What is the value of z?

    Answer: z=
    numeric
    2 attempts remaining
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In a two-way table, the numbers in each column have to add up to the total for that column. In the same way, the numbers in each row have to add up to the total for that row. Looking at the second column:

    5+z=y5+z=56z=51

    Submit your answer as:

ID is: 1648 Seed is: 669

Contingency tables

Learners in Grade 11 were interviewed about their favourite animal and the results of the survey were recorded in the following two-way table:

Dog Cat Total
Girls 57 24 z
Boys 33 59 y
Total x 83 173
  1. What is the value of x?

    Answer: x=
    numeric
    2 attempts remaining
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In a two-way table, the numbers in each column have to add up to the total for that column. In the same way, the numbers in each row have to add up to the total for that row. Looking at the first column:

    57+33=xx=90

    Submit your answer as:
  2. What is the value of y?

    Answer: y=
    numeric
    2 attempts remaining
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In a two-way table, the numbers in each column have to add up to the total for that column. In the same way, the numbers in each row have to add up to the total for that row. Looking at the second row:

    33+59=yy=92

    Submit your answer as:
  3. What is the value of z?

    Answer: z=
    numeric
    2 attempts remaining
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    In a two-way table, the numbers in each column have to add up to the total for that column. In the same way, the numbers in each row have to add up to the total for that row. Looking at the third column:

    z+y=173z+92=173z=81

    Submit your answer as:

ID is: 2291 Seed is: 7021

Tree diagrams with replacement

A box contains 2 orange sweets and 10 blue sweets. A sweet is drawn at random and then replaced. Another sweet is then taken from the box and replaced.

The following tree diagram has been constructed using the information above.

Determine the values of the missing probabilities: B, Q, and U.

INSTRUCTION: Your answer must be a fraction in its lowest terms.
Answer:
  1. B=
  2. Q=
  3. U=
fraction
fraction
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The probability P of any event is calculated as follows:

P=favourable outcomesNumber of possible outcomesTotal number of 

You are told a sweet picked from the box is replaced before the second sweet is picked. This means that the numbers of favourable and the total number of possible outcomes are fixed.

Use this information to find the values of the missing probabilities.


STEP: Determine the values of the quantities B, Q, and U
[−3 points ⇒ 0 / 3 points left]
NOTE: Tree diagrams allow us to see all the possible outcomes of an event. Each branch in a tree diagram represents a possible outcome. We calculate the probability of an outcome using probabilities from different branches.

To construct the tree diagram, we need to calculate two probabilities:

  • the probability of picking a orange sweet, and
  • the probability of picking a blue sweet.

We calculate the probability P of any event as follows:

P=favourable outcomesNumber of possible outcomesTotal number of 

The question says that a sweet picked from the box is replaced before picking the second sweet. This means that the numbers of favourable and the total possible outcomes are fixed.

The probability of picking a orange sweet is:

P=Number of orange sweetsTotal number of sweets=212=16

The probability of picking a blue sweet is:

P=Number of orange sweetsTotal number of sweets=1012=56

Using these results, we can fill in the missing probabilities in the tree diagram. The completed tree diagram is:

The values of the quantities B, Q, and U are:

  1. B=56
  2. Q=56
  3. U=16

Submit your answer as: andand

ID is: 2291 Seed is: 5597

Tree diagrams with replacement

A box contains 7 black sweets and 2 blue sweets. A sweet is drawn at random and then replaced. Another sweet is then taken from the box and replaced.

The following tree diagram has been constructed using the information above.

Determine the values of the missing probabilities: O, W, and F.

INSTRUCTION: Your answer must be a fraction in its lowest terms.
Answer:
  1. O=
  2. W=
  3. F=
fraction
fraction
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The probability P of any event is calculated as follows:

P=favourable outcomesNumber of possible outcomesTotal number of 

You are told a sweet picked from the box is replaced before the second sweet is picked. This means that the numbers of favourable and the total number of possible outcomes are fixed.

Use this information to find the values of the missing probabilities.


STEP: Determine the values of the quantities O, W, and F
[−3 points ⇒ 0 / 3 points left]
NOTE: Tree diagrams allow us to see all the possible outcomes of an event. Each branch in a tree diagram represents a possible outcome. We calculate the probability of an outcome using probabilities from different branches.

To construct the tree diagram, we need to calculate two probabilities:

  • the probability of picking a black sweet, and
  • the probability of picking a blue sweet.

We calculate the probability P of any event as follows:

P=favourable outcomesNumber of possible outcomesTotal number of 

The question says that a sweet picked from the box is replaced before picking the second sweet. This means that the numbers of favourable and the total possible outcomes are fixed.

The probability of picking a black sweet is:

P=Number of black sweetsTotal number of sweets=79

The probability of picking a blue sweet is:

P=Number of black sweetsTotal number of sweets=29

Using these results, we can fill in the missing probabilities in the tree diagram. The completed tree diagram is:

The values of the quantities O, W, and F are:

  1. O=29
  2. W=29
  3. F=29

Submit your answer as: andand

ID is: 2291 Seed is: 982

Tree diagrams with replacement

A bag contains 10 orange balls and 4 brown balls. Rethabile picks a ball at random from the bag and replaces it back in the bag. He mixes the balls in the bag and then picks another ball at random from the bag.

The following tree diagram has been constructed using the information above.

Determine the values of the missing probabilities: B, G, and A.

INSTRUCTION: Your answer must be a fraction in its lowest terms.
Answer:
  1. B=
  2. G=
  3. A=
fraction
fraction
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The probability P of any event is calculated as follows:

P=favourable outcomesNumber of possible outcomesTotal number of 

You are told a ball picked from the bag is replaced before the second ball is picked. This means that the numbers of favourable and the total number of possible outcomes are fixed.

Use this information to find the values of the missing probabilities.


STEP: Determine the values of the quantities B, G, and A
[−3 points ⇒ 0 / 3 points left]
NOTE: Tree diagrams allow us to see all the possible outcomes of an event. Each branch in a tree diagram represents a possible outcome. We calculate the probability of an outcome using probabilities from different branches.

To construct the tree diagram, we need to calculate two probabilities:

  • the probability of picking a orange ball, and
  • the probability of picking a brown ball.

We calculate the probability P of any event as follows:

P=favourable outcomesNumber of possible outcomesTotal number of 

The question says that a ball picked from the bag is replaced before picking the second ball. This means that the numbers of favourable and the total possible outcomes are fixed.

The probability of picking a orange ball is:

P=Number of orange ballsTotal number of balls=1014=57

The probability of picking a brown ball is:

P=Number of orange ballsTotal number of balls=414=27

Using these results, we can fill in the missing probabilities in the tree diagram. The completed tree diagram is:

The values of the quantities B, G, and A are:

  1. B=27
  2. G=57
  3. A=27

Submit your answer as: andand

ID is: 2297 Seed is: 5556

Tree diagrams

There are only yellow marbles and red marbles in a bag. There are 5 yellow marbles and 7 red marbles. Jidda takes at random a marble from the bag. He puts the marble back in the bag. He then takes at random another marble from the bag.

What is the probability of not choosing a red marble from the bag?

TIP: You need to construct a tree diagram and use it to calculate the probability for this event.
INSTRUCTION: Your answer must be a fraction in its lowest terms.
Answer:

The probability of not choosing a red marble from the bag is .

fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start by constructing a tree diagram using the information given in the question. Then select the branches that produce the desired outcome. Use the selected branches to work out the probability.


STEP: Construct the tree diagram
[−2 points ⇒ 2 / 4 points left]

We need to calculate the probability of not choosing a red marble from the bag. Before we can calculate, we need to construct a tree diagram.

NOTE: Tree diagrams allow us to see all the possible outcomes of an event. Each branch in a tree diagram represents a possible outcome. We calculate the probability of an outcome using probabilities from different branches.

To construct the tree diagram, we need to calculate two probabilities:

  • the probability of picking a yellow marble, and
  • the probability of picking a red marble.

The question says that a marble picked from the bag is replaced before picking the second marble. This means that the numbers of favourable and the total possible outcomes are fixed.

The probability of picking a yellow marble is:

P=Number of yellow marblesTotal number of marbles=512

The probability of picking a red marble is:

P=Number of yellow marblesTotal number of marbles=712

Using these results, we can now construct a tree diagram for the events described:

NOTE: The branch with double lines is the one that produces the desired outcome.

STEP: Calculate the probability of not choosing a red marble from the bag
[−2 points ⇒ 0 / 4 points left]

To get the probability of the desired outcome, we need to multiply along the selected branch.

P, the probability of not choosing a red marble from the bag is:

P=512×512=25144

The probability of not choosing a red marble from the bag is 25144.


Submit your answer as:

ID is: 2297 Seed is: 5586

Tree diagrams

A bag contains 6 black balls and 8 brown balls. Adeboye picks a ball at random from the bag and replaces it back in the bag. He mixes the balls in the bag and then picks another ball at random from the bag.

What is the probability of not choosing a black ball from the bag?

TIP: You need to construct a tree diagram and use it to calculate the probability for this event.
INSTRUCTION: Your answer must be a fraction in its lowest terms.
Answer:

The probability of not choosing a black ball from the bag is .

fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start by constructing a tree diagram using the information given in the question. Then select the branches that produce the desired outcome. Use the selected branches to work out the probability.


STEP: Construct the tree diagram
[−2 points ⇒ 2 / 4 points left]

We need to calculate the probability of not choosing a black ball from the bag. Before we can calculate, we need to construct a tree diagram.

NOTE: Tree diagrams allow us to see all the possible outcomes of an event. Each branch in a tree diagram represents a possible outcome. We calculate the probability of an outcome using probabilities from different branches.

To construct the tree diagram, we need to calculate two probabilities:

  • the probability of picking a black ball, and
  • the probability of picking a brown ball.

The question says that a ball picked from the bag is replaced before picking the second ball. This means that the numbers of favourable and the total possible outcomes are fixed.

The probability of picking a black ball is:

P=Number of black ballsTotal number of balls=614=37

The probability of picking a brown ball is:

P=Number of black ballsTotal number of balls=814=47

Using these results, we can now construct a tree diagram for the events described:

NOTE: The branch with double lines is the one that produces the desired outcome.

STEP: Calculate the probability of not choosing a black ball from the bag
[−2 points ⇒ 0 / 4 points left]

To get the probability of the desired outcome, we need to multiply along the selected branch.

P, the probability of not choosing a black ball from the bag is:

P=47×47=1649

The probability of not choosing a black ball from the bag is 1649.


Submit your answer as:

ID is: 2297 Seed is: 1300

Tree diagrams

A box contains 3 blue sweets and 5 brown sweets. A sweet is drawn at random and then replaced. Another sweet is then taken from the box and replaced.

What is the probability of first choosing a blue sweet before a brown sweet from the box?

TIP: You need to construct a tree diagram and use it to calculate the probability for this event.
INSTRUCTION: Your answer must be a fraction in its lowest terms.
Answer:

The probability of first choosing a blue sweet before a brown sweet from the box is .

fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start by constructing a tree diagram using the information given in the question. Then select the branches that produce the desired outcome. Use the selected branches to work out the probability.


STEP: Construct the tree diagram
[−2 points ⇒ 2 / 4 points left]

We need to calculate the probability of first choosing a blue sweet before a brown sweet from the box. Before we can calculate, we need to construct a tree diagram.

NOTE: Tree diagrams allow us to see all the possible outcomes of an event. Each branch in a tree diagram represents a possible outcome. We calculate the probability of an outcome using probabilities from different branches.

To construct the tree diagram, we need to calculate two probabilities:

  • the probability of picking a blue sweet, and
  • the probability of picking a brown sweet.

The question says that a sweet picked from the box is replaced before picking the second sweet. This means that the numbers of favourable and the total possible outcomes are fixed.

The probability of picking a blue sweet is:

P=Number of blue sweetsTotal number of sweets=38

The probability of picking a brown sweet is:

P=Number of blue sweetsTotal number of sweets=58

Using these results, we can now construct a tree diagram for the events described:

NOTE: The branch with double lines is the one that produces the desired outcome.

STEP: Calculate the probability of first choosing a blue sweet before a brown sweet from the box
[−2 points ⇒ 0 / 4 points left]

To get the probability of the desired outcome, we need to multiply along the selected branch.

P, the probability of first choosing a blue sweet before a brown sweet from the box is:

P=38×58=1564

The probability of first choosing a blue sweet before a brown sweet from the box is 1564.


Submit your answer as:

ID is: 4341 Seed is: 2970

Contingency tables: comparing probabilities

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

The table below shows data on the monthly income of employed people in two residential areas. Representative samples were used in the collection of the data.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 366 40 406
3,200x<25,600 732 1,018 1,750
x25,600 1,342 569 1,911
Total 2,440 1,627 4,067

Consider the following events:

  • A: a randomly chosen person from Area 1 earns at least N=25,600.
  • B: a randomly chosen person from Area 2 earns at least N=25,600.

Which event is more likely? Show calculations to support your answer.

INSTRUCTION: Round your calculated values to two decimal places.
Answer:
  • P(A)=
  • P(B)=

Therefore, it is more likely that:

one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The probability of an event is given by the formula

P(E)=favourable outcomesnumber ofpossible outcomestotal

Choose the correct values from the table.


STEP: Calculate the probability that a randomly chosen person from Area 1 earns at least N=25,600 per month
[−1 point ⇒ 2 / 3 points left]

The probability of an event is given by the formula

P(E)=favourable outcomesnumber ofpossible outcomestotal

For this question, we are looking at each area separately.

So, the total possible outcomes is the total number of people in Area 1.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 366 40 406
3,200x<25,600 732 1,018 1,750
x25,600 1,342 569 1,911
Total 2,440 1,627 4,067

The probability that a randomly chosen person from Area 1 earns at least N=25,600 per month:

P(A)=1,3422,440=1120=0.55

STEP: Calculate the probability that a randomly chosen person from Area 2 earns at least N=25,600 per month
[−2 points ⇒ 0 / 3 points left]
MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 366 40 406
3,200x<25,600 732 1,018 1,750
x25,600 1,342 569 1,911
Total 2,440 1,627 4,067

The probability that a randomly chosen person from Area 2 earns at least N=25,600 per month:

P(B)=5691,627=0.34972...0.35

Therefore it is more likely that a person from Area 1 earns at least N=25,600 per month.


Submit your answer as: andand

ID is: 4341 Seed is: 5721

Contingency tables: comparing probabilities

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

The table below shows data on the monthly income of employed people in two residential areas. Representative samples were used in the collection of the data.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 1,029 411 1,440
3,200x<25,600 515 550 1,065
x25,600 514 411 925
Total 2,058 1,372 3,430

Consider the following events:

  • A: a randomly chosen person from Area 1 earns less than N=3,200.
  • B: a randomly chosen person from Area 2 earns less than N=3,200.

Which event is more likely? Show calculations to support your answer.

INSTRUCTION: Round your calculated values to two decimal places.
Answer:
  • P(A)=
  • P(B)=

Therefore, it is more likely that:

one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The probability of an event is given by the formula

P(E)=favourable outcomesnumber ofpossible outcomestotal

Choose the correct values from the table.


STEP: Calculate the probability that a randomly chosen person from Area 1 earns less than N=3,200 per month
[−1 point ⇒ 2 / 3 points left]

The probability of an event is given by the formula

P(E)=favourable outcomesnumber ofpossible outcomestotal

For this question, we are looking at each area separately.

So, the total possible outcomes is the total number of people in Area 1.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 1,029 411 1,440
3,200x<25,600 515 550 1,065
x25,600 514 411 925
Total 2,058 1,372 3,430

The probability that a randomly chosen person from Area 1 earns less than N=3,200 per month:

P(A)=1,0292,058=12=0.5

STEP: Calculate the probability that a randomly chosen person from Area 2 earns less than N=3,200 per month
[−2 points ⇒ 0 / 3 points left]
MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 1,029 411 1,440
3,200x<25,600 515 550 1,065
x25,600 514 411 925
Total 2,058 1,372 3,430

The probability that a randomly chosen person from Area 2 earns less than N=3,200 per month:

P(B)=4111,372=0.29956...0.3

Therefore it is more likely that a person from Area 1 earns less than N=3,200 per month.


Submit your answer as: andand

ID is: 4341 Seed is: 6467

Contingency tables: comparing probabilities

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

The table below shows data on the monthly income of employed people in two residential areas. Representative samples were used in the collection of the data.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 22 104 126
3,200x<25,600 247 1,048 1,295
x25,600 627 941 1,568
Total 896 2,093 2,989

Consider the following events:

  • A: a randomly chosen person from Area 1 earns at least N=25,600.
  • B: a randomly chosen person from Area 2 earns at least N=25,600.

Which event is more likely? Show calculations to support your answer.

INSTRUCTION: Round your calculated values to two decimal places.
Answer:
  • P(A)=
  • P(B)=

Therefore, it is more likely that:

one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The probability of an event is given by the formula

P(E)=favourable outcomesnumber ofpossible outcomestotal

Choose the correct values from the table.


STEP: Calculate the probability that a randomly chosen person from Area 1 earns at least N=25,600 per month
[−1 point ⇒ 2 / 3 points left]

The probability of an event is given by the formula

P(E)=favourable outcomesnumber ofpossible outcomestotal

For this question, we are looking at each area separately.

So, the total possible outcomes is the total number of people in Area 1.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 22 104 126
3,200x<25,600 247 1,048 1,295
x25,600 627 941 1,568
Total 896 2,093 2,989

The probability that a randomly chosen person from Area 1 earns at least N=25,600 per month:

P(A)=627896=0.69977...0.7

STEP: Calculate the probability that a randomly chosen person from Area 2 earns at least N=25,600 per month
[−2 points ⇒ 0 / 3 points left]
MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 22 104 126
3,200x<25,600 247 1,048 1,295
x25,600 627 941 1,568
Total 896 2,093 2,989

The probability that a randomly chosen person from Area 2 earns at least N=25,600 per month:

P(B)=9412,093=0.44959...0.45

Therefore it is more likely that a person from Area 1 earns at least N=25,600 per month.


Submit your answer as: andand

ID is: 1647 Seed is: 6903

Contingency tables of given events

During a trial 73 females and 117 males volunteered to participate for research purposes. The results indicated that 12 females tested negative. The total amount of 120 tested positive during the trial.

Complete the contingency table with the given information, and calculate the missing values (A-H).

Answer:
Females Males Total
Positive A: B: C:
Not Positive D: E: F:
Total G: H: Total:
numeric
numeric
numeric
numeric
numeric
numeric
numeric
numeric
numeric
2 attempts remaining
STEP: <no title>
[−9 points ⇒ 0 / 9 points left]

Always first try and substitute the information that was given in the question into your table. Once you have substituted these values you will do some easy addition and subtraction calculations to find the other missing values.

Here the value of 73 (Females) and 117 (Males) are given and you add them to find the total.

73+117=190

Then you can calculate the value of A:

7312=61

Try doing the same for the other values and check them in the table below:

Females Males Total
Positive A= 61 B= 59 C= 120
Not Positive D= 12 E= 58 F= 70
Total G= 73 H= 117 Total= 190

Submit your answer as: andandandandandandandand

ID is: 1647 Seed is: 9220

Contingency tables of given events

There are 200 students in Year 9 and 56 of them have a packed lunch. In Year 6 there are 131 students and 51 of them do not have a packed lunch.

Complete the contingency table with the given information, and calculate the missing values (A-H).

Answer:
Year 9 Year 6 Total
a packed lunch A: B: C:
Not a packed lunch D: E: F:
Total G: H: Total:
numeric
numeric
numeric
numeric
numeric
numeric
numeric
numeric
numeric
2 attempts remaining
STEP: <no title>
[−9 points ⇒ 0 / 9 points left]

Always first try and substitute the information that was given in the question into your table. Once you have substituted these values you will do some easy addition and subtraction calculations to find the other missing values.

Here the value of 200 (Year 9) and 131 (Year 6) are given and you add them to find the total.

200+131=331

Then you can calculate the value of A:

200144=56

Try doing the same for the other values and check them in the table below:

Year 9 Year 6 Total
a packed lunch A= 56 B= 80 C= 136
Not a packed lunch D= 144 E= 51 F= 195
Total G= 200 H= 131 Total= 331

Submit your answer as: andandandandandandandand

ID is: 1647 Seed is: 1816

Contingency tables of given events

There are 78 students in Year 8 and 7 of them have Science. In Year 11 there are 69 students and 63 of them do not have Science.

Complete the contingency table with the given information, and calculate the missing values (A-H).

Answer:
Year 8 Year 11 Total
Science A: B: C:
Not Science D: E: F:
Total G: H: Total:
numeric
numeric
numeric
numeric
numeric
numeric
numeric
numeric
numeric
2 attempts remaining
STEP: <no title>
[−9 points ⇒ 0 / 9 points left]

Always first try and substitute the information that was given in the question into your table. Once you have substituted these values you will do some easy addition and subtraction calculations to find the other missing values.

Here the value of 78 (Year 8) and 69 (Year 11) are given and you add them to find the total.

78+69=147

Then you can calculate the value of A:

7871=7

Try doing the same for the other values and check them in the table below:

Year 8 Year 11 Total
Science A= 7 B= 6 C= 13
Not Science D= 71 E= 63 F= 134
Total G= 78 H= 69 Total= 147

Submit your answer as: andandandandandandandand

ID is: 1645 Seed is: 9508

Tree diagrams without replacement

There are only red marbles and yellow marbles in a bag. There are 11 red marbles and 3 yellow marbles. Ifetayo takes a marble from the bag at random. She does not put the marble back in the bag. She then takes at random another marble from the bag.

The following tree diagram has been constructed using the information above.

Determine the values of the missing probabilities: L, C, and B.

INSTRUCTION: Give your answer as a simplified fraction.
Answer:
  1. L=
  2. C=
  3. B=
fraction
fraction
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question says that the marbles which are removed are not replaced. This means that the total number of possible outcomes is not fixed, and that the outcome of the first event will affect the outcome of the second event.


STEP: Determine the values of L, C, and B
[−3 points ⇒ 0 / 3 points left]

Tree diagrams allow us to see all the possible outcomes of an event or sequence of events. Each branch in the diagram represents a possible outcome.

In this question, we have two choices, and two options for each choice. This means there are four possible final outcomes, each of which has some probability. The value shown on each of the branches represents the probability of making that choice. We need to determine the missing probability values.

Let's look at the option where we first pick a red marble. There are 14 marbles in total in the bag, and 11 of these are red, so the probability of picking a red marble is 1114.

Similarly, since there are 3 yellow marbles in the bag, the probability of the first pick being a yellow marble is 314.

Before we make our second pick, we need to remember that the first marble is not replaced in the bag before the second marble is chosen. This is called 'sampling without replacement'. When we sample without replacement, the result of out second pick depends on the result of our first pick.

The total number of possible outcomes for the second pick is now 13, since there is one less marble in the bag (the marble we took out in the first pick).

If we picked a red marble on our first pick, we now have two options again. There would be 10 red marbles left in the bag, since we have already removed one, and there are a total of 13 marbles in the bag. So the probability of our second pick being red if the first pick was red is 1013. The number of yellow marbles in the bag will not change if we removed a red marble the first time, so the probability of picking a yellow marble after removing a red marble is 313.

Similarly, if we picked a yellow marble the first time, there will be 2 yellow marbles and 11 red marbles left out of a total of 13 marbles. So the probability of picking a yellow marble after removing a yellow marble is 213. And the probability of picking a red marble after removing a yellow marble is 1113.

Using these results, we can fill in the missing probabilities on the tree diagram. The completed tree diagram is:

The values of L, C, and B are:

  1. L=1114
  2. C=314
  3. B=313

Submit your answer as: andand

ID is: 1645 Seed is: 9436

Tree diagrams without replacement

A bag contains 3 black balls and 12 yellow balls. Nkosingiphile picks a ball at random from the bag and does not replace it in the bag. He mixes the balls in the bag and then picks another ball at random from the bag.

The following tree diagram has been constructed using the information above.

Determine the values of the missing probabilities: T, U, and X.

INSTRUCTION: Give your answer as a simplified fraction.
Answer:
  1. T=
  2. U=
  3. X=
fraction
fraction
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question says that the balls which are removed are not replaced. This means that the total number of possible outcomes is not fixed, and that the outcome of the first event will affect the outcome of the second event.


STEP: Determine the values of T, U, and X
[−3 points ⇒ 0 / 3 points left]

Tree diagrams allow us to see all the possible outcomes of an event or sequence of events. Each branch in the diagram represents a possible outcome.

In this question, we have two choices, and two options for each choice. This means there are four possible final outcomes, each of which has some probability. The value shown on each of the branches represents the probability of making that choice. We need to determine the missing probability values.

Let's look at the option where we first pick a black ball. There are 15 balls in total in the bag, and 3 of these are black, so the probability of picking a black ball is 15.

Similarly, since there are 12 yellow balls in the bag, the probability of the first pick being a yellow ball is 45.

Before we make our second pick, we need to remember that the first ball is not replaced in the bag before the second ball is chosen. This is called 'sampling without replacement'. When we sample without replacement, the result of out second pick depends on the result of our first pick.

The total number of possible outcomes for the second pick is now 14, since there is one less ball in the bag (the ball we took out in the first pick).

If we picked a black ball on our first pick, we now have two options again. There would be 2 black balls left in the bag, since we have already removed one, and there are a total of 14 balls in the bag. So the probability of our second pick being black if the first pick was black is 17. The number of yellow balls in the bag will not change if we removed a black ball the first time, so the probability of picking a yellow ball after removing a black ball is 67.

Similarly, if we picked a yellow ball the first time, there will be 11 yellow balls and 3 black balls left out of a total of 14 balls. So the probability of picking a yellow ball after removing a yellow ball is 1114. And the probability of picking a black ball after removing a yellow ball is 314.

Using these results, we can fill in the missing probabilities on the tree diagram. The completed tree diagram is:

The values of T, U, and X are:

  1. T=45
  2. U=17
  3. X=67

Submit your answer as: andand

ID is: 1645 Seed is: 1303

Tree diagrams without replacement

A box contains 12 blue sweets and 2 white sweets. A sweet is drawn at random, and is not replaced. Another sweet is then taken from the box and is also not replaced.

The following tree diagram has been constructed using the information above.

Determine the values of the missing probabilities: W, G, and A.

INSTRUCTION: Give your answer as a simplified fraction.
Answer:
  1. W=
  2. G=
  3. A=
fraction
fraction
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The question says that the sweets which are removed are not replaced. This means that the total number of possible outcomes is not fixed, and that the outcome of the first event will affect the outcome of the second event.


STEP: Determine the values of W, G, and A
[−3 points ⇒ 0 / 3 points left]

Tree diagrams allow us to see all the possible outcomes of an event or sequence of events. Each branch in the diagram represents a possible outcome.

In this question, we have two choices, and two options for each choice. This means there are four possible final outcomes, each of which has some probability. The value shown on each of the branches represents the probability of making that choice. We need to determine the missing probability values.

Let's look at the option where we first pick a blue sweet. There are 14 sweets in total in the box, and 12 of these are blue, so the probability of picking a blue sweet is 67.

Similarly, since there are 2 white sweets in the box, the probability of the first pick being a white sweet is 17.

Before we make our second pick, we need to remember that the first sweet is not replaced in the box before the second sweet is chosen. This is called 'sampling without replacement'. When we sample without replacement, the result of out second pick depends on the result of our first pick.

The total number of possible outcomes for the second pick is now 13, since there is one less sweet in the box (the sweet we took out in the first pick).

If we picked a blue sweet on our first pick, we now have two options again. There would be 11 blue sweets left in the box, since we have already removed one, and there are a total of 13 sweets in the box. So the probability of our second pick being blue if the first pick was blue is 1113. The number of white sweets in the box will not change if we removed a blue sweet the first time, so the probability of picking a white sweet after removing a blue sweet is 213.

Similarly, if we picked a white sweet the first time, there will be 1 white sweets and 12 blue sweets left out of a total of 13 sweets. So the probability of picking a white sweet after removing a white sweet is 113. And the probability of picking a blue sweet after removing a white sweet is 1213.

Using these results, we can fill in the missing probabilities on the tree diagram. The completed tree diagram is:

The values of W, G, and A are:

  1. W=213
  2. G=1213
  3. A=113

Submit your answer as: andand

ID is: 3873 Seed is: 7388

Two-way tables

Adapted from DBE Nov 2016 Grade 12, P1, Q11
Maths formulas

A survey was conducted among 60 boys and 120 girls to determine how many of them watched a movie in the period the examinations were written. Their responses are shown in the partially completed table below.

WATCHED A MOVIE DURING EXAMS DID NOT WATCH A MOVIE DURING EXAMS TOTALS
Male a 33
Female 48 72
Totals 75 b 180
  1. Calculate the values of a and b.

    Answer:
    1. a=
    2. b=
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    In a frequency table:

    • the totals of each row appear at the right.
    • the totals of each column appear at the bottom.
    • sum of the row totals equals the sum of the column totals.

    Use this information to find the values of the missing quantities.


    STEP: Solve for x
    [−2 points ⇒ 0 / 2 points left]

    In a frequency table:

    • the totals of each row appear at the right.
    • the totals of each column appear at the bottom.
    • sum of the row totals equals the sum of the column totals.

    Calculate the missing value a:

    a+48=75a=7548a=27

    Calculate the missing value b:

    33+72=b105=b

    The completed table is:

    WATCHED A MOVIE DURING EXAMS DID NOT WATCH A MOVIE DURING EXAMS TOTALS
    Male 27 33 60
    Female 48 72 120
    Totals 75 105 180

    Submit your answer as: and
  2. Are the events "being male" and "did not watch a movie during examinations" mutually exclusive? Give a reason for your answer.

    Answer: The events "being male" and "did not watch a movie during examinations" are because the probability of "being male" and "did not watch a movie during examinations" is . 2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Two events are called mutually exclusive if they cannot occur at the same time. You may find it helpful to revise mutually exclusive events in the Everything Maths textbook.


    STEP: Determine the answer
    [−2 points ⇒ 0 / 2 points left]

    Two events are "mutually exclusive" if they cannot occur at the same time. In other words, the probality of the two events happening at the same time is zero.

    From the table, there are 33 students who are "male" and "did not watch a movie during examinations". Since the number of students for these two events is not zero, it means the events occurred at the same time.

    NOTE: We can use the probability of the two events to determine the answer. The probability of "being male" and "did not watch a movie during examinations" is 33180. This probability is not zero. This means it is possible for both events to occur at the same time.

    The answer is: The events "being male" and "did not watch a movie during examinations" are not mutually exclusive because the the probability of "being male" and "did not watch a movie during examinations" is not equal to zero.


    Submit your answer as: and
  3. If a learner who participated in this survey is chosen at random, what is the probability that the learner:

    1. Watched a movie in the period during which the examinations were written?
    2. Is not a female and did not watch a movie in the period during which the examinations were written?
    INSTRUCTION: Give your answers as simplified fractions.
    Answer:
    1. The probability that the learner watched a movie in the period during which the examinations were written is .
    2. The probability that the learner is not a female and did not watch a movie in the period during which the examinations were written is .
    one-of
    type(numeric.abserror(0.01))
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the information provided in the two-way table to calculate the probabilities.

    Remember, the probability formula is:

    P=favourable outcomesNumber ofpossible outcomesTotal number of

    STEP: Use the information in the two-way table to calculate the probability for part 1
    [−2 points ⇒ 2 / 4 points left]

    The probability P that the learner watched a movie in the period during which the examinations were written is

    P=who watched a movieNumber of studentsstudentsTotal number of=27+48180=75180=512

    STEP: Use the information in the two-way table to calculate the probability for part 2
    [−2 points ⇒ 0 / 4 points left]

    The questions asks for the probability that the learner is not a female and did not watch a movie in the period during which the examinations were written.

    TIP: A learner who is not a female is a male. In other words, we need to calculate the probability that a male learner did not watch a movie in the period during which the examinations were written.
    P=who did not watch TVNumber of male studentsstudentsTotal number of =33180=1160

    Submit your answer as: and

ID is: 3873 Seed is: 9347

Two-way tables

Adapted from DBE Nov 2016 Grade 12, P1, Q11
Maths formulas

A survey was conducted among 80 boys and 100 girls to determine how many of them played video games in the period the examinations were written. Their responses are shown in the partially completed table below.

PLAYED VIDEO GAMES DURING EXAMS DID NOT PLAY VIDEO GAMES DURING EXAMS TOTALS
Male 61 19
Female a 33
Totals 128 b 180
  1. Calculate the values of a and b.

    Answer:
    1. a=
    2. b=
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    In a frequency table:

    • the totals of each row appear at the right.
    • the totals of each column appear at the bottom.
    • sum of the row totals equals the sum of the column totals.

    Use this information to find the values of the missing quantities.


    STEP: Solve for x
    [−2 points ⇒ 0 / 2 points left]

    In a frequency table:

    • the totals of each row appear at the right.
    • the totals of each column appear at the bottom.
    • sum of the row totals equals the sum of the column totals.

    Calculate the missing value a:

    61+a=128a=12861a=67

    Calculate the missing value b:

    19+33=b52=b

    The completed table is:

    PLAYED VIDEO GAMES DURING EXAMS DID NOT PLAY VIDEO GAMES DURING EXAMS TOTALS
    Male 61 19 80
    Female 67 33 100
    Totals 128 52 180

    Submit your answer as: and
  2. Are the events "being female" and "did not play video games during examinations" mutually exclusive? Give a reason for your answer.

    Answer: The events "being female" and "did not play video games during examinations" are because the probability of "being female" and "did not play video games during examinations" is . 2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Two events are called mutually exclusive if they cannot occur at the same time. You may find it helpful to revise mutually exclusive events in the Everything Maths textbook.


    STEP: Determine the answer
    [−2 points ⇒ 0 / 2 points left]

    Two events are "mutually exclusive" if they cannot occur at the same time. In other words, the probality of the two events happening at the same time is zero.

    From the table, there are 33 students who are "female" and "did not play video games during examinations". Since the number of students for these two events is not zero, it means the events occurred at the same time.

    NOTE: We can use the probability of the two events to determine the answer. The probability of "being female" and "did not play video games during examinations" is 33180. This probability is not zero. This means it is possible for both events to occur at the same time.

    The answer is: The events "being female" and "did not play video games during examinations" are not mutually exclusive because the the probability of "being female" and "did not play video games during examinations" is not equal to zero.


    Submit your answer as: and
  3. If a learner who participated in this survey is chosen at random, what is the probability that the learner:

    1. Played video games in the period during which the examinations were written?
    2. Is not a male and did not play video games in the period during which the examinations were written?
    INSTRUCTION: Give your answers as simplified fractions.
    Answer:
    1. The probability that the learner played video games in the period during which the examinations were written is .
    2. The probability that the learner is not a male and did not play video games in the period during which the examinations were written is .
    one-of
    type(numeric.abserror(0.01))
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the information provided in the two-way table to calculate the probabilities.

    Remember, the probability formula is:

    P=favourable outcomesNumber ofpossible outcomesTotal number of

    STEP: Use the information in the two-way table to calculate the probability for part 1
    [−2 points ⇒ 2 / 4 points left]

    The probability P that the learner played video games in the period during which the examinations were written is

    P=who played video gamesNumber of studentsstudentsTotal number of=61+67180=128180=3245

    STEP: Use the information in the two-way table to calculate the probability for part 2
    [−2 points ⇒ 0 / 4 points left]

    The questions asks for the probability that the learner is not a male and did not play video games in the period during which the examinations were written.

    TIP: A learner who is not a male is a female. In other words, we need to calculate the probability that a female learner did not play video games in the period during which the examinations were written.
    P=who did not watch TVNumber of female studentsstudentsTotal number of =33180=1160

    Submit your answer as: and

ID is: 3873 Seed is: 6990

Two-way tables

Adapted from DBE Nov 2016 Grade 12, P1, Q11
Maths formulas

A survey was conducted among 60 boys and 120 girls to determine how many of them watched a movie in the period the examinations were written. Their responses are shown in the partially completed table below.

WATCHED A MOVIE DURING EXAMS DID NOT WATCH A MOVIE DURING EXAMS TOTALS
Male 33 27
Female 100 m
Totals n 47 180
  1. Calculate the values of m and n.

    Answer:
    1. m=
    2. n=
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    In a frequency table:

    • the totals of each row appear at the right.
    • the totals of each column appear at the bottom.
    • sum of the row totals equals the sum of the column totals.

    Use this information to find the values of the missing quantities.


    STEP: Solve for x
    [−2 points ⇒ 0 / 2 points left]

    In a frequency table:

    • the totals of each row appear at the right.
    • the totals of each column appear at the bottom.
    • sum of the row totals equals the sum of the column totals.

    Calculate the missing value m:

    27+m=47m=4727m=20

    Calculate the missing value n:

    33+100=n133=n

    The completed table is:

    WATCHED A MOVIE DURING EXAMS DID NOT WATCH A MOVIE DURING EXAMS TOTALS
    Male 33 27 60
    Female 100 20 120
    Totals 133 47 180

    Submit your answer as: and
  2. Are the events "being male" and "did not watch a movie during examinations" mutually exclusive? Give a reason for your answer.

    Answer: The events "being male" and "did not watch a movie during examinations" are because the probability of "being male" and "did not watch a movie during examinations" is . 2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Two events are called mutually exclusive if they cannot occur at the same time. You may find it helpful to revise mutually exclusive events in the Everything Maths textbook.


    STEP: Determine the answer
    [−2 points ⇒ 0 / 2 points left]

    Two events are "mutually exclusive" if they cannot occur at the same time. In other words, the probality of the two events happening at the same time is zero.

    From the table, there are 27 students who are "male" and "did not watch a movie during examinations". Since the number of students for these two events is not zero, it means the events occurred at the same time.

    NOTE: We can use the probability of the two events to determine the answer. The probability of "being male" and "did not watch a movie during examinations" is 27180. This probability is not zero. This means it is possible for both events to occur at the same time.

    The answer is: The events "being male" and "did not watch a movie during examinations" are not mutually exclusive because the the probability of "being male" and "did not watch a movie during examinations" is not equal to zero.


    Submit your answer as: and
  3. If a learner who participated in this survey is chosen at random, what is the probability that the learner:

    1. Watched a movie in the period during which the examinations were written?
    2. Is not a male and did not watch a movie in the period during which the examinations were written?
    INSTRUCTION: Give your answers as simplified fractions.
    Answer:
    1. The probability that the learner watched a movie in the period during which the examinations were written is .
    2. The probability that the learner is not a male and did not watch a movie in the period during which the examinations were written is .
    one-of
    type(numeric.abserror(0.01))
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the information provided in the two-way table to calculate the probabilities.

    Remember, the probability formula is:

    P=favourable outcomesNumber ofpossible outcomesTotal number of

    STEP: Use the information in the two-way table to calculate the probability for part 1
    [−2 points ⇒ 2 / 4 points left]

    The probability P that the learner watched a movie in the period during which the examinations were written is

    P=who watched a movieNumber of studentsstudentsTotal number of=33+100180=133180

    STEP: Use the information in the two-way table to calculate the probability for part 2
    [−2 points ⇒ 0 / 4 points left]

    The questions asks for the probability that the learner is not a male and did not watch a movie in the period during which the examinations were written.

    TIP: A learner who is not a male is a female. In other words, we need to calculate the probability that a female learner did not watch a movie in the period during which the examinations were written.
    P=who did not watch TVNumber of female studentsstudentsTotal number of =20180=19

    Submit your answer as: and

ID is: 3972 Seed is: 6333

Probability: sampling with replacement

Adapted from DBE Nov 2015 Grade 12, P1, Q11.3
Maths formulas

There are v red sweets and 2 green sweets in a box. Chizoba randomly selects one sweet from the box, records his choice, and returns the sweet to the box. He then randomly selects a second sweet from the box, records his choice, and returns it to the box. It is known that the probability that Chizoba will select two sweets of different colours from the box is 48%.

Calculate how many red sweets are in the box.

Answer: There are red sweets.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by drawing and labelling a tree diagram for this scenario.


STEP: Draw a tree diagram and label it
[−2 points ⇒ 4 / 6 points left]

Since we are given a scenario where there are two consecutive events taking place, it would be useful to draw a tree diagram to show us all of the possible outcomes.

There are a total of v+2 sweets in the box. v of these are red, and 2 are green. So the probability of randomly selecting a red sweet is

P(R)=vv+2

and the probability of selecting a green sweet is

P(G)=2v+2

Since the first sweet is replaced in the box before the second sweet is selected, the probability of selecting a red or green sweet is the same for the second draw as it is for the first draw. We can now draw and label a tree diagram for these events:


STEP: Work out the probability of getting two sweets of the same colour
[−2 points ⇒ 2 / 6 points left]

We are told in the question that the probability of drawing two sweets of different colours is 48%. This is the same as saying that the probability of drawing two sweets of the same colour is 10048=52%. This means that the total probability of getting either two red or two green sweets is 52100.

P(R,R)+P(G,G)=52100

Substituting and simplifying:

P(R,R)=(vv+2)(vv+2)=v2(v+2)2
P(G,G)=(2v+2)(2v+2)=4(v+2)2

So

v2(v+2)2+4(v+2)2=52100

We can use this equation to find the value of v.


STEP: Simplify and solve for v
[−2 points ⇒ 0 / 6 points left]

We can now simplify and rearrange to find the value of v.

v2+4(v+2)2=52100
100(v2+4)=52(v2+4v+4)
48v2208v+192=03v213v+12=0(3v4)(v3)=0
v=43 or v=3

Since v represents the number of red sweets in the box, v must be a positive natural number. So there are 3 red sweets in the box.


Submit your answer as:

ID is: 3972 Seed is: 9708

Probability: sampling with replacement

Adapted from DBE Nov 2015 Grade 12, P1, Q11.3
Maths formulas

There are s green balls and 2 blue balls in a box. Jabulile randomly selects one ball from the box, records her choice, and returns the ball to the box. She then randomly selects a second ball from the box, records her choice, and returns it to the box. It is known that the probability that Jabulile will select two balls of the same colour from the box is 52%.

Calculate how many green balls are in the box.

Answer: There are green balls.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by drawing and labelling a tree diagram for this scenario.


STEP: Draw a tree diagram and label it
[−2 points ⇒ 4 / 6 points left]

Since we are given a scenario where there are two consecutive events taking place, it would be useful to draw a tree diagram to show us all of the possible outcomes.

There are a total of s+2 balls in the box. s of these are green, and 2 are blue. So the probability of randomly selecting a green ball is

P(G)=ss+2

and the probability of selecting a blue ball is

P(B)=2s+2

Since the first ball is replaced in the box before the second ball is selected, the probability of selecting a green or blue ball is the same for the second draw as it is for the first draw. We can now draw and label a tree diagram for these events:


STEP: Work out the probability of getting two balls of the same colour
[−2 points ⇒ 2 / 6 points left]

We are told in the question that the probability of drawing two balls of the same colour is 52%. This means that the total probability of getting either two green or two blue balls is 52100.

P(G,G)+P(B,B)=52100

Substituting and simplifying:

P(G,G)=(ss+2)(ss+2)=s2(s+2)2
P(B,B)=(2s+2)(2s+2)=4(s+2)2

So

s2(s+2)2+4(s+2)2=52100

We can use this equation to find the value of s.


STEP: Simplify and solve for s
[−2 points ⇒ 0 / 6 points left]

We can now simplify and rearrange to find the value of s.

s2+4(s+2)2=52100
100(s2+4)=52(s2+4s+4)
48s2208s+192=03s213s+12=0(3s4)(s3)=0
s=43 or s=3

Since s represents the number of green balls in the box, s must be a positive natural number. So there are 3 green balls in the box.


Submit your answer as:

ID is: 3972 Seed is: 9673

Probability: sampling with replacement

Adapted from DBE Nov 2015 Grade 12, P1, Q11.3
Maths formulas

There are c green sweets and 2 red sweets in a box. Abdoul randomly selects one sweet from the box, records his choice, and returns the sweet to the box. He then randomly selects a second sweet from the box, records his choice, and returns it to the box. It is known that the probability that Abdoul will select two sweets of different colours from the box is 48%.

Calculate how many green sweets are in the box.

Answer: There are green sweets.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by drawing and labelling a tree diagram for this scenario.


STEP: Draw a tree diagram and label it
[−2 points ⇒ 4 / 6 points left]

Since we are given a scenario where there are two consecutive events taking place, it would be useful to draw a tree diagram to show us all of the possible outcomes.

There are a total of c+2 sweets in the box. c of these are green, and 2 are red. So the probability of randomly selecting a green sweet is

P(G)=cc+2

and the probability of selecting a red sweet is

P(R)=2c+2

Since the first sweet is replaced in the box before the second sweet is selected, the probability of selecting a green or red sweet is the same for the second draw as it is for the first draw. We can now draw and label a tree diagram for these events:


STEP: Work out the probability of getting two sweets of the same colour
[−2 points ⇒ 2 / 6 points left]

We are told in the question that the probability of drawing two sweets of different colours is 48%. This is the same as saying that the probability of drawing two sweets of the same colour is 10048=52%. This means that the total probability of getting either two green or two red sweets is 52100.

P(G,G)+P(R,R)=52100

Substituting and simplifying:

P(G,G)=(cc+2)(cc+2)=c2(c+2)2
P(R,R)=(2c+2)(2c+2)=4(c+2)2

So

c2(c+2)2+4(c+2)2=52100

We can use this equation to find the value of c.


STEP: Simplify and solve for c
[−2 points ⇒ 0 / 6 points left]

We can now simplify and rearrange to find the value of c.

c2+4(c+2)2=52100
100(c2+4)=52(c2+4c+4)
48c2208c+192=03c213c+12=0(3c4)(c3)=0
c=43 or c=3

Since c represents the number of green sweets in the box, c must be a positive natural number. So there are 3 green sweets in the box.


Submit your answer as:

ID is: 1643 Seed is: 1383

Using tree diagrams: application problems

There are only red marbles and brown marbles in a bag. There are 7 red marbles and 3 brown marbles. Karabou takes a marble from the bag at random. She does not put the marble back in the bag. She then takes another marble from the bag at random.

What is the probability of choosing two red marbles from the bag?

INSTRUCTION: Give your answer as a simplified fraction.
Answer: The probability of choosing two red marbles from the bag is .
fraction
2 attempts remaining
HINT: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by drawing a tree diagram. Select the sequence of branches that produces the desired outcome, and use them to work out the probability.


STEP: Determine the probability of the desired outcome
[−3 points ⇒ 0 / 3 points left]

Tree diagrams allow us to see all the possible outcomes of an event or sequence of events. Each branch in a tree diagram represents a possible outcome. We can calculate the probability of a specific outcome by multiplying the probability values on each branch leading to that outcome.

Let's start by working out the probabilities for the first choice. There are 7 red marbles and 3 brown marbles, giving a total of 10 marbles. So the probability of picking a red marble is 710, and the probability of picking a brown marble is 310.

Before we make our second pick, we need to remember that the first marble is not replaced in the bag before the second marble is chosen. When we sample without replacement, our first pick affects the results of the second pick.

The total number of possible outcomes for the second pick is now 9, since there is one less marble in the bag (the marble we took out in the first pick).

If we picked a red marble on our first pick, there would be 6 red marbles left in the bag, since we have already removed one, and there are a total of 9 marbles in the bag. So the probability of our second pick being red if the first pick was red is 23. The number of brown marbles in the bag will not change if we removed a red marble the first time, so the probability of picking a brown marble after removing a red marble is 13.

Similarly, if we picked a brown marble the first time, there will be 2 brown marbles and 7 red marbles left out of a total of 9 marbles. So the probability of picking a brown marble after removing a brown marble is 29. And the probability of picking a red marble after removing a brown marble is 79.

Using these results, we can fill in the missing probabilities on the tree diagram. The completed tree diagram is:

The probability of choosing two red marbles from the bag is the product of the probabilities along the branches that lead to the desired outcome.

P, the probability of choosing two red marbles from the bag, is:

P=710×23=715

The probability of choosing two red marbles from the bag is 715.


Submit your answer as:

ID is: 1643 Seed is: 1869

Using tree diagrams: application problems

A box contains 8 red sweets and 4 green sweets. A sweet is drawn at random, and is not replaced. Another sweet is then taken from the box and is also not replaced.

What is the probability of not choosing a green sweet from the box?

INSTRUCTION: Give your answer as a simplified fraction.
Answer: The probability of not choosing a green sweet from the box is .
fraction
2 attempts remaining
HINT: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by drawing a tree diagram. Select the sequence of branches that produces the desired outcome, and use them to work out the probability.


STEP: Determine the probability of the desired outcome
[−3 points ⇒ 0 / 3 points left]

Tree diagrams allow us to see all the possible outcomes of an event or sequence of events. Each branch in a tree diagram represents a possible outcome. We can calculate the probability of a specific outcome by multiplying the probability values on each branch leading to that outcome.

Let's start by working out the probabilities for the first choice. There are 8 red sweets and 4 green sweets, giving a total of 12 sweets. So the probability of picking a red sweet is 23, and the probability of picking a green sweet is 13.

Before we make our second pick, we need to remember that the first sweet is not replaced in the box before the second sweet is chosen. When we sample without replacement, our first pick affects the results of the second pick.

The total number of possible outcomes for the second pick is now 11, since there is one less sweet in the box (the sweet we took out in the first pick).

If we picked a red sweet on our first pick, there would be 7 red sweets left in the box, since we have already removed one, and there are a total of 11 sweets in the box. So the probability of our second pick being red if the first pick was red is 711. The number of green sweets in the box will not change if we removed a red sweet the first time, so the probability of picking a green sweet after removing a red sweet is 411.

Similarly, if we picked a green sweet the first time, there will be 3 green sweets and 8 red sweets left out of a total of 11 sweets. So the probability of picking a green sweet after removing a green sweet is 311. And the probability of picking a red sweet after removing a green sweet is 811.

Using these results, we can fill in the missing probabilities on the tree diagram. The completed tree diagram is:

The probability of not choosing a green sweet from the box is the product of the probabilities along the branches that lead to the desired outcome.

P, the probability of not choosing a green sweet from the box, is:

P=23×711=1433

The probability of not choosing a green sweet from the box is 1433.


Submit your answer as:

ID is: 1643 Seed is: 3010

Using tree diagrams: application problems

A box contains 5 blue sweets and 5 black sweets. A sweet is drawn at random, and is not replaced. Another sweet is then taken from the box and is also not replaced.

What is the probability of first choosing a blue sweet and then choosing a black sweet from the box?

INSTRUCTION: Give your answer as a simplified fraction.
Answer: The probability of first choosing a blue sweet and then choosing a black sweet from the box is .
fraction
2 attempts remaining
HINT: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by drawing a tree diagram. Select the sequence of branches that produces the desired outcome, and use them to work out the probability.


STEP: Determine the probability of the desired outcome
[−3 points ⇒ 0 / 3 points left]

Tree diagrams allow us to see all the possible outcomes of an event or sequence of events. Each branch in a tree diagram represents a possible outcome. We can calculate the probability of a specific outcome by multiplying the probability values on each branch leading to that outcome.

Let's start by working out the probabilities for the first choice. There are 5 blue sweets and 5 black sweets, giving a total of 10 sweets. So the probability of picking a blue sweet is 12, and the probability of picking a black sweet is 12.

Before we make our second pick, we need to remember that the first sweet is not replaced in the box before the second sweet is chosen. When we sample without replacement, our first pick affects the results of the second pick.

The total number of possible outcomes for the second pick is now 9, since there is one less sweet in the box (the sweet we took out in the first pick).

If we picked a blue sweet on our first pick, there would be 4 blue sweets left in the box, since we have already removed one, and there are a total of 9 sweets in the box. So the probability of our second pick being blue if the first pick was blue is 49. The number of black sweets in the box will not change if we removed a blue sweet the first time, so the probability of picking a black sweet after removing a blue sweet is 59.

Similarly, if we picked a black sweet the first time, there will be 4 black sweets and 5 blue sweets left out of a total of 9 sweets. So the probability of picking a black sweet after removing a black sweet is 49. And the probability of picking a blue sweet after removing a black sweet is 59.

Using these results, we can fill in the missing probabilities on the tree diagram. The completed tree diagram is:

The probability of first choosing a blue sweet and then choosing a black sweet from the box is the product of the probabilities along the branches that lead to the desired outcome.

P, the probability of first choosing a blue sweet and then choosing a black sweet from the box, is:

P=12×59=518

The probability of first choosing a blue sweet and then choosing a black sweet from the box is 518.


Submit your answer as:

ID is: 1646 Seed is: 916

Using tree diagrams to calculate probabilities with replacement

A box contains 12 brown sweets and 2 green sweets. A sweet is drawn at random and then replaced. Another sweet is then taken from the box and replaced.

What is the probability of first choosing a green sweet an then choosing a brown sweet from the box?

INSTRUCTION: Give you answer as a simplified fraction.
Answer: The probability of first choosing a green sweet an then choosing a brown sweet from the box is .
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by drawing a tree diagram. Select the sequence of branches that produces the desired outcome, and use it to work out the probability.


STEP: Determine the probability of the desired outcome
[−3 points ⇒ 0 / 3 points left]

Tree diagrams allow us to see all the possible outcomes of an event or sequence of events. Each branch in a tree diagram represents a possible outcome. We can calculate the probability of a specific outcome by multiplying the probability values on each branch leading to that outcome.

To construct a tree diagram for this question, we need to calculate both the probability of picking a brown sweet, and the probability of picking a green sweet.

We calculate the probability of an event as follows:

P=favourable outcomesNumber ofpossible outcomesTotal number of

The question tells us that the first sweet picked from the box is replaced before the second sweet is picked. This means that the number of favourable outcomes, and the total number of possible outcomes, are both fixed, since we are always picking from the same set of sweets.

The probability of picking a brown sweet is:

P(brown)=brown sweetsNumber ofof sweetsTotal number=1214=67

The probability of picking a green sweet is:

P(green)=green sweetsNumber ofof sweetsTotal number=214=17

Using these results, we can construct the tree diagram. The completed diagram is:

The probability of first choosing a green sweet an then choosing a brown sweet from the box is the product of the probabilities along the branches that lead to the desired outcome.

P, the probability of first choosing a green sweet an then choosing a brown sweet from the box, is:

P=17×67=649

The probability of first choosing a green sweet an then choosing a brown sweet from the box is 649.


Submit your answer as:

ID is: 1646 Seed is: 9008

Using tree diagrams to calculate probabilities with replacement

A bag contains 10 yellow balls and 4 red balls. Adebankole picks a ball at random from the bag and replaces it back in the bag. He mixes the balls in the bag and then picks another ball at random from the bag.

What is the probability of choosing two red balls from the bag?

INSTRUCTION: Give you answer as a simplified fraction.
Answer: The probability of choosing two red balls from the bag is .
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by drawing a tree diagram. Select the sequence of branches that produces the desired outcome, and use it to work out the probability.


STEP: Determine the probability of the desired outcome
[−3 points ⇒ 0 / 3 points left]

Tree diagrams allow us to see all the possible outcomes of an event or sequence of events. Each branch in a tree diagram represents a possible outcome. We can calculate the probability of a specific outcome by multiplying the probability values on each branch leading to that outcome.

To construct a tree diagram for this question, we need to calculate both the probability of picking a yellow ball, and the probability of picking a red ball.

We calculate the probability of an event as follows:

P=favourable outcomesNumber ofpossible outcomesTotal number of

The question tells us that the first ball picked from the bag is replaced before the second ball is picked. This means that the number of favourable outcomes, and the total number of possible outcomes, are both fixed, since we are always picking from the same set of balls.

The probability of picking a yellow ball is:

P(yellow)=yellow ballsNumber ofof ballsTotal number=1014=57

The probability of picking a red ball is:

P(red)=red ballsNumber ofof ballsTotal number=414=27

Using these results, we can construct the tree diagram. The completed diagram is:

The probability of choosing two red balls from the bag is the product of the probabilities along the branches that lead to the desired outcome.

P, the probability of choosing two red balls from the bag, is:

P=27×27=449

The probability of choosing two red balls from the bag is 449.


Submit your answer as:

ID is: 1646 Seed is: 7770

Using tree diagrams to calculate probabilities with replacement

There are only black marbles and green marbles in a bag. There are 12 black marbles and 2 green marbles. Zainab takes at random a marble from the bag. She puts the marble back in the bag. She then takes another marble from the bag at random.

What is the probability of not choosing a green marble from the bag?

INSTRUCTION: Give you answer as a simplified fraction.
Answer: The probability of not choosing a green marble from the bag is .
fraction
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by drawing a tree diagram. Select the sequence of branches that produces the desired outcome, and use it to work out the probability.


STEP: Determine the probability of the desired outcome
[−3 points ⇒ 0 / 3 points left]

Tree diagrams allow us to see all the possible outcomes of an event or sequence of events. Each branch in a tree diagram represents a possible outcome. We can calculate the probability of a specific outcome by multiplying the probability values on each branch leading to that outcome.

To construct a tree diagram for this question, we need to calculate both the probability of picking a black marble, and the probability of picking a green marble.

We calculate the probability of an event as follows:

P=favourable outcomesNumber ofpossible outcomesTotal number of

The question tells us that the first marble picked from the bag is replaced before the second marble is picked. This means that the number of favourable outcomes, and the total number of possible outcomes, are both fixed, since we are always picking from the same set of marbles.

The probability of picking a black marble is:

P(black)=black marblesNumber ofof marblesTotal number=1214=67

The probability of picking a green marble is:

P(green)=green marblesNumber ofof marblesTotal number=214=17

Using these results, we can construct the tree diagram. The completed diagram is:

The probability of not choosing a green marble from the bag is the product of the probabilities along the branches that lead to the desired outcome.

P, the probability of not choosing a green marble from the bag, is:

P=67×67=3649

The probability of not choosing a green marble from the bag is 3649.


Submit your answer as:

ID is: 4343 Seed is: 1663

Contingency tables: calculating probabilities

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

The table below shows data on the monthly income of employed people in two residential areas. Representative samples were used in the collection of the data.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 924 756 1,680
3,200x<25,600 1,651 1,189 2,840
x25,600 66 216 282
Total 2,641 2,161 4,802

Use the table to answer the following questions.

INSTRUCTION: Give your answers as simplified fractions, or rounded to two decimal places.
Answer:

What is the probability that a person chosen randomly from the entire sample will be:

  1. A person who earns at least N=3,200 but less than N=25,600 per month?
  2. A person from Area 1 who earns N=25,600 or more per month?
  3. A person from Area 2 who earns at least N=3,200 per month?
one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The probability of an event is given by the formula

P(E)=favourable outcomesnumber ofpossible outcomestotal

Choose the correct values from the table.


STEP: 1. What is the probability that a randomly chosen person will earn between N=3,200 and N=25,600?
[−2 points ⇒ 3 / 5 points left]

The probability of an event is given by

P(E)=favourable outcomesnumber ofpossible outcomestotal

The number of favourable outcomes is the total number of people earning between N=3,200 and N=25,600, and the number of possible outcomes is the grand total.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 924 756 1,680
3,200x<25,600 1,651 1,189 2,840
x25,600 66 216 282
Total 2,641 2,161 4,802

The probability that a randomly chosen person earns between N=3,200 and N=25,600:

2,8404,802=1,4202,401=0.59142...0.59

STEP: 2. What is the probability that a randomly chosen person will be a person from Area 1 who earns N=25,600 or more per month?
[−1 point ⇒ 2 / 5 points left]
MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 924 756 1,680
3,200x<25,600 1,651 1,189 2,840
x25,600 66 216 282
Total 2,641 2,161 4,802

The favourable outcome is the number of people who live in Area 1 and earn N=25,600 or more per month, and the number of possible outcomes is the grand total.

he probability that a randomly chosen person is a person from Area 1 who earns N=25,600 or more per month:

664,802=332,401=0.01374...0.01

STEP: 3. What is the probability that a randomly chosen person is a person from Area 2 who earns at least N=3,200 per month?
[−2 points ⇒ 0 / 5 points left]
MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 924 756 1,680
3,200x<25,600 1,651 1,189 2,840
x25,600 66 216 282
Total 2,641 2,161 4,802

The favourable outcome is the number of people who live in Area 2 and earn at least N=3,200, and the number of possible outcomes is the grand total.

The probability that a randomly chosen person is a person from Area 2 who earns at least N=3,200 per month:

1,189+2164,802=0.29258...0.29

Submit your answer as: andand

ID is: 4343 Seed is: 4127

Contingency tables: calculating probabilities

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

The table below shows data on the monthly income of employed people in two residential areas. Representative samples were used in the collection of the data.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 1,002 1,224 2,226
3,200x<25,600 1,003 818 1,821
x25,600 222 680 902
Total 2,227 2,722 4,949

Use the table to answer the following questions.

INSTRUCTION: Give your answers as simplified fractions, or rounded to two decimal places.
Answer:

What is the probability that a person chosen randomly from the entire sample will be:

  1. A person who earns less than N=3,200 per month?
  2. A person from Area 2 who earns at least N=3,200 but less than N=25,600 per month?
  3. A person from Area 1 who earns less than N=25,600 per month?
one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The probability of an event is given by the formula

P(E)=favourable outcomesnumber ofpossible outcomestotal

Choose the correct values from the table.


STEP: 1. What is the probability that a randomly chosen person will earn less than N=3,200?
[−2 points ⇒ 3 / 5 points left]

The probability of an event is given by

P(E)=favourable outcomesnumber ofpossible outcomestotal

The number of favourable outcomes is the total number of people earning less than N=3,200, and the number of possible outcomes is the grand total.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 1,002 1,224 2,226
3,200x<25,600 1,003 818 1,821
x25,600 222 680 902
Total 2,227 2,722 4,949

The probability that a randomly chosen person earns less than N=3,200:

2,2264,949=318707=0.44978...0.45

STEP: 2. What is the probability that a randomly chosen person will be a person from Area 2 who earns at least N=3,200 but less than N=25,600 per month?
[−1 point ⇒ 2 / 5 points left]
MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 1,002 1,224 2,226
3,200x<25,600 1,003 818 1,821
x25,600 222 680 902
Total 2,227 2,722 4,949

The favourable outcome is the number of people who live in Area 2 and earn at least N=3,200 but less than N=25,600 per month, and the number of possible outcomes is the grand total.

he probability that a randomly chosen person is a person from Area 2 who earns at least N=3,200 but less than N=25,600 per month:

8184,949=0.16528...0.17

STEP: 3. What is the probability that a randomly chosen person is a person from Area 1 who earns less than N=25,600 per month?
[−2 points ⇒ 0 / 5 points left]
MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 1,002 1,224 2,226
3,200x<25,600 1,003 818 1,821
x25,600 222 680 902
Total 2,227 2,722 4,949

The favourable outcome is the number of people who live in Area 1 and earn less than N=25,600, and the number of possible outcomes is the grand total.

The probability that a randomly chosen person is a person from Area 1 who earns less than N=25,600 per month:

1,002+1,0034,949=0.40513...0.41

Submit your answer as: andand

ID is: 4343 Seed is: 7877

Contingency tables: calculating probabilities

Adapted from DBE Nov 2015 Grade 11, P1, Q9
Maths formulas

The table below shows data on the monthly income of employed people in two residential areas. Representative samples were used in the collection of the data.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 888 159 1,047
3,200x<25,600 1,038 879 1,917
x25,600 1,036 557 1,593
Total 2,962 1,595 4,557

Use the table to answer the following questions.

INSTRUCTION: Give your answers as simplified fractions, or rounded to two decimal places.
Answer:

What is the probability that a person chosen randomly from the entire sample will be:

  1. From Area 2?
  2. A person from Area 1 who earns at least N=3,200 but less than N=25,600 per month?
  3. A person from Area 1 who earns less than N=25,600 per month?
one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
one-of
type(numeric.abserror(0.01))
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The probability of an event is given by the formula

P(E)=favourable outcomesnumber ofpossible outcomestotal

Choose the correct values from the table.


STEP: 1. What is the probability that a randomly chosen person will be from Area 2?
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The probability of an event is given by

P(E)=favourable outcomesnumber ofpossible outcomestotal

The number of favourable outcomes is the total number of people in Area 2, and the number of possible outcomes is the grand total.

MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 888 159 1,047
3,200x<25,600 1,038 879 1,917
x25,600 1,036 557 1,593
Total 2,962 1,595 4,557

The probability that a randomly chosen person is from Area 2:

1,5954,557=0.35001...0.35

STEP: 2. What is the probability that a randomly chosen person will be a person from Area 1 who earns at least N=3,200 but less than N=25,600 per month?
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MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 888 159 1,047
3,200x<25,600 1,038 879 1,917
x25,600 1,036 557 1,593
Total 2,962 1,595 4,557

The favourable outcome is the number of people who live in Area 1 and earn at least N=3,200 but less than N=25,600 per month, and the number of possible outcomes is the grand total.

he probability that a randomly chosen person is a person from Area 1 who earns at least N=3,200 but less than N=25,600 per month:

1,0384,557=3461,519=0.22778...0.23

STEP: 3. What is the probability that a randomly chosen person is a person from Area 1 who earns less than N=25,600 per month?
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MONTHLY INCOME
(IN RANDS)
Area 1 Area 2 TOTAL
x<3,200 888 159 1,047
3,200x<25,600 1,038 879 1,917
x25,600 1,036 557 1,593
Total 2,962 1,595 4,557

The favourable outcome is the number of people who live in Area 1 and earn less than N=25,600, and the number of possible outcomes is the grand total.

The probability that a randomly chosen person is a person from Area 1 who earns less than N=25,600 per month:

888+1,0384,557=6421,519=0.42264...0.42

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